Results in $mathbb{R}^{2n}$ applying to $mathbb{C}^{n}$











up vote
0
down vote

favorite












So I managed to show that in the vectorspace $mathbb{R}^{2n}$ boundedness of a set is equivalent to total boundedness. "Intuitively" I would say that this result applies aswell to $mathbb{C}^{n}$. The problem now is that, although I know that you can view $mathbb{C}$ as a $mathbb{R}$ vectorspace, what bothers me is, what happens if we view $mathbb{C}$ as a $mathbb{C}$ vectorspace ?



And more general to what extend do results in $mathbb{R}^{2n}$ apply to $mathbb{C}^{n}$ ?










share|cite|improve this question




















  • 2




    The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
    – Xander Henderson
    Nov 21 at 13:47










  • Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
    – Christian Singer
    Nov 21 at 13:49

















up vote
0
down vote

favorite












So I managed to show that in the vectorspace $mathbb{R}^{2n}$ boundedness of a set is equivalent to total boundedness. "Intuitively" I would say that this result applies aswell to $mathbb{C}^{n}$. The problem now is that, although I know that you can view $mathbb{C}$ as a $mathbb{R}$ vectorspace, what bothers me is, what happens if we view $mathbb{C}$ as a $mathbb{C}$ vectorspace ?



And more general to what extend do results in $mathbb{R}^{2n}$ apply to $mathbb{C}^{n}$ ?










share|cite|improve this question




















  • 2




    The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
    – Xander Henderson
    Nov 21 at 13:47










  • Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
    – Christian Singer
    Nov 21 at 13:49















up vote
0
down vote

favorite









up vote
0
down vote

favorite











So I managed to show that in the vectorspace $mathbb{R}^{2n}$ boundedness of a set is equivalent to total boundedness. "Intuitively" I would say that this result applies aswell to $mathbb{C}^{n}$. The problem now is that, although I know that you can view $mathbb{C}$ as a $mathbb{R}$ vectorspace, what bothers me is, what happens if we view $mathbb{C}$ as a $mathbb{C}$ vectorspace ?



And more general to what extend do results in $mathbb{R}^{2n}$ apply to $mathbb{C}^{n}$ ?










share|cite|improve this question















So I managed to show that in the vectorspace $mathbb{R}^{2n}$ boundedness of a set is equivalent to total boundedness. "Intuitively" I would say that this result applies aswell to $mathbb{C}^{n}$. The problem now is that, although I know that you can view $mathbb{C}$ as a $mathbb{R}$ vectorspace, what bothers me is, what happens if we view $mathbb{C}$ as a $mathbb{C}$ vectorspace ?



And more general to what extend do results in $mathbb{R}^{2n}$ apply to $mathbb{C}^{n}$ ?







linear-algebra complex-numbers vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 13:53

























asked Nov 21 at 13:38









Christian Singer

344113




344113








  • 2




    The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
    – Xander Henderson
    Nov 21 at 13:47










  • Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
    – Christian Singer
    Nov 21 at 13:49
















  • 2




    The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
    – Xander Henderson
    Nov 21 at 13:47










  • Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
    – Christian Singer
    Nov 21 at 13:49










2




2




The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
– Xander Henderson
Nov 21 at 13:47




The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
– Xander Henderson
Nov 21 at 13:47












Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
– Christian Singer
Nov 21 at 13:49






Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
– Christian Singer
Nov 21 at 13:49












1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Your basic question could be rephrased in terms of structure. Specifically, "What structures can we impose on $mathbb{R}^{2n}$ and $mathbb{C}^n$ such that the two objects are indistinguishable? What structures would distinguish these two objects?" There are a ton of structures running around here. For example:




  • As sets, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are indistinguishable: there is (at least one) bijective function between the two spaces. In particular, the map $f: mathbb{R}^2 to mathbb{C}$ given by $f(x,y) = x+iy$ is bijective, and we can go from $mathbb{R}^{2n}$ to $mathbb{C}^n$ by applying $f$ componentwise. This implies that the two sets have the same cardinality, which, from the point of view of sets, means that the two objects are the same. Therefore any set theoretic property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$.


  • As topological spaces, $mathbb{R}^{2n}$ and $mathbb{C}^n$. Indeed, this can be understood by seeing that as metric spaces the two objects are the same: in $mathbb{R}^{2n}$ the metric is induced by the norm
    $$ | (x_1, x_2, dotsc x_{2n}) | = left( sum_{j=1}^{2n} x_j^2 right)^{1/2}, $$
    and in $mathbb{C}^n$ the metric is induced by
    $$ | (z_1, z_2, dotsc, z_n) |
    = left( sum_{j=1}^{n} |z_j|^2 right)^{1/2}, $$

    where $|z| = sqrt{Re(z)^2 + Im(z)^2}$ is the modulus of $z$. It is not too difficult to verify that the map
    $$(x_1,x_2,dotsc,x_{2n-1},x_{2n}) mapsto (x_1+ix_2, dotsc, x_{2n-1}+ix_{2n})$$
    is an isometry, which means that the two spaces are isometric, i.e. they are indistinguishable as metric spaces. Since the usual topologies on both spaces are induced by the metrics, this also means that the two spaces are homeomorphic as topological spaces, which means that they are indistinguishable in that context, as well.



    Therefore any metric or topological property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$. As an example, you have shown that boundedness implies total boundedness in $mathbb{R}^{2n}$, which automatically implies that the same result holds for $mathbb{C}^n$. Similarly, $mathbb{R}^{2n}$ has the Heine-Borel property (a set is compact if and only if it is closed and bounded), which implies that $mathbb{C}^n$ also has this property.



    Do note, however, that we could impose strange metrics or topologies on either space, and we would get something different. For example, the discrete metric on $mathbb{R}$, given by
    $$ d(x,y) = begin{cases} 0 & text{if $x=y$, and} \ 1 & text{otherwise,} end{cases} $$
    turns $mathbb{R}$ into a completely different space.




  • As vector spaces, things get a little more interesting. If we regard both $mathbb{R}^{2n}$ and $mathbb{C}^{n}$ as $mathbb{R}$-vector spaces, then they are indistinguishable. Indeed, there is an "obvious" map sending ${langle 1,0rangle, langle 0,1rangle}$ (a basis for $mathbb{R}^2$) and ${1,i}$ (a basis for $mathbb{C}$ as an $mathbb{R}$-vector space), hence the two $mathbb{R}$-vector spaces are both two dimensional vector spaces over $mathbb{R}$ and therefore isomorphic.



    On the other hand, it is pretty unusual to view $mathbb{C}$ as an $mathbb{R}$-vector space–when we do this, we lose all of the interesting and useful algebraic structure which can be found in the field $mathbb{C}$. Hence it is much more common to view $mathbb{C}^n$ as a $mathbb{C}$-vector space (indeed, if we are working with vector spaces and we aren't doing this, then why would we even bother with the notation $mathbb{C}^n$?). In this setting, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are quite different spaces, owing to the differnet structures of $mathbb{C}$ and $mathbb{R}$. Hence there are things that are true about $mathbb{R}^{2n}$ which are not true about $mathbb{C}^n$, and vice versa. As a simple example, with $n=1$, $mathbb{C}$ is an algebraically closed field, while $mathbb{R}^{2}$ is not even a field.




More generally, your question touches on a branch of mathematics called category theory. I am not very conversant in category theory, so I don't want to say too much about this, but the basic goal of this theory is to study structures in mathematics in an abstract way. If we can show that two objects are isomorphic in some category, then any property which holds for one object will also hold for the other (in that category). Hence if you want to know which properties of $mathbb{R}^{2n}$ easily extend to $mathbb{C}^n$, a good place to start might be to understand the categories in which the two objects are identical, i.e. which structures can be imposed on the two objects which fail to distinguish between them.






share|cite|improve this answer





















  • Thanks alot for this indepth response! I appreciate that!
    – Christian Singer
    Nov 21 at 15:16











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007735%2fresults-in-mathbbr2n-applying-to-mathbbcn%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Your basic question could be rephrased in terms of structure. Specifically, "What structures can we impose on $mathbb{R}^{2n}$ and $mathbb{C}^n$ such that the two objects are indistinguishable? What structures would distinguish these two objects?" There are a ton of structures running around here. For example:




  • As sets, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are indistinguishable: there is (at least one) bijective function between the two spaces. In particular, the map $f: mathbb{R}^2 to mathbb{C}$ given by $f(x,y) = x+iy$ is bijective, and we can go from $mathbb{R}^{2n}$ to $mathbb{C}^n$ by applying $f$ componentwise. This implies that the two sets have the same cardinality, which, from the point of view of sets, means that the two objects are the same. Therefore any set theoretic property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$.


  • As topological spaces, $mathbb{R}^{2n}$ and $mathbb{C}^n$. Indeed, this can be understood by seeing that as metric spaces the two objects are the same: in $mathbb{R}^{2n}$ the metric is induced by the norm
    $$ | (x_1, x_2, dotsc x_{2n}) | = left( sum_{j=1}^{2n} x_j^2 right)^{1/2}, $$
    and in $mathbb{C}^n$ the metric is induced by
    $$ | (z_1, z_2, dotsc, z_n) |
    = left( sum_{j=1}^{n} |z_j|^2 right)^{1/2}, $$

    where $|z| = sqrt{Re(z)^2 + Im(z)^2}$ is the modulus of $z$. It is not too difficult to verify that the map
    $$(x_1,x_2,dotsc,x_{2n-1},x_{2n}) mapsto (x_1+ix_2, dotsc, x_{2n-1}+ix_{2n})$$
    is an isometry, which means that the two spaces are isometric, i.e. they are indistinguishable as metric spaces. Since the usual topologies on both spaces are induced by the metrics, this also means that the two spaces are homeomorphic as topological spaces, which means that they are indistinguishable in that context, as well.



    Therefore any metric or topological property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$. As an example, you have shown that boundedness implies total boundedness in $mathbb{R}^{2n}$, which automatically implies that the same result holds for $mathbb{C}^n$. Similarly, $mathbb{R}^{2n}$ has the Heine-Borel property (a set is compact if and only if it is closed and bounded), which implies that $mathbb{C}^n$ also has this property.



    Do note, however, that we could impose strange metrics or topologies on either space, and we would get something different. For example, the discrete metric on $mathbb{R}$, given by
    $$ d(x,y) = begin{cases} 0 & text{if $x=y$, and} \ 1 & text{otherwise,} end{cases} $$
    turns $mathbb{R}$ into a completely different space.




  • As vector spaces, things get a little more interesting. If we regard both $mathbb{R}^{2n}$ and $mathbb{C}^{n}$ as $mathbb{R}$-vector spaces, then they are indistinguishable. Indeed, there is an "obvious" map sending ${langle 1,0rangle, langle 0,1rangle}$ (a basis for $mathbb{R}^2$) and ${1,i}$ (a basis for $mathbb{C}$ as an $mathbb{R}$-vector space), hence the two $mathbb{R}$-vector spaces are both two dimensional vector spaces over $mathbb{R}$ and therefore isomorphic.



    On the other hand, it is pretty unusual to view $mathbb{C}$ as an $mathbb{R}$-vector space–when we do this, we lose all of the interesting and useful algebraic structure which can be found in the field $mathbb{C}$. Hence it is much more common to view $mathbb{C}^n$ as a $mathbb{C}$-vector space (indeed, if we are working with vector spaces and we aren't doing this, then why would we even bother with the notation $mathbb{C}^n$?). In this setting, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are quite different spaces, owing to the differnet structures of $mathbb{C}$ and $mathbb{R}$. Hence there are things that are true about $mathbb{R}^{2n}$ which are not true about $mathbb{C}^n$, and vice versa. As a simple example, with $n=1$, $mathbb{C}$ is an algebraically closed field, while $mathbb{R}^{2}$ is not even a field.




More generally, your question touches on a branch of mathematics called category theory. I am not very conversant in category theory, so I don't want to say too much about this, but the basic goal of this theory is to study structures in mathematics in an abstract way. If we can show that two objects are isomorphic in some category, then any property which holds for one object will also hold for the other (in that category). Hence if you want to know which properties of $mathbb{R}^{2n}$ easily extend to $mathbb{C}^n$, a good place to start might be to understand the categories in which the two objects are identical, i.e. which structures can be imposed on the two objects which fail to distinguish between them.






share|cite|improve this answer





















  • Thanks alot for this indepth response! I appreciate that!
    – Christian Singer
    Nov 21 at 15:16















up vote
1
down vote



accepted










Your basic question could be rephrased in terms of structure. Specifically, "What structures can we impose on $mathbb{R}^{2n}$ and $mathbb{C}^n$ such that the two objects are indistinguishable? What structures would distinguish these two objects?" There are a ton of structures running around here. For example:




  • As sets, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are indistinguishable: there is (at least one) bijective function between the two spaces. In particular, the map $f: mathbb{R}^2 to mathbb{C}$ given by $f(x,y) = x+iy$ is bijective, and we can go from $mathbb{R}^{2n}$ to $mathbb{C}^n$ by applying $f$ componentwise. This implies that the two sets have the same cardinality, which, from the point of view of sets, means that the two objects are the same. Therefore any set theoretic property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$.


  • As topological spaces, $mathbb{R}^{2n}$ and $mathbb{C}^n$. Indeed, this can be understood by seeing that as metric spaces the two objects are the same: in $mathbb{R}^{2n}$ the metric is induced by the norm
    $$ | (x_1, x_2, dotsc x_{2n}) | = left( sum_{j=1}^{2n} x_j^2 right)^{1/2}, $$
    and in $mathbb{C}^n$ the metric is induced by
    $$ | (z_1, z_2, dotsc, z_n) |
    = left( sum_{j=1}^{n} |z_j|^2 right)^{1/2}, $$

    where $|z| = sqrt{Re(z)^2 + Im(z)^2}$ is the modulus of $z$. It is not too difficult to verify that the map
    $$(x_1,x_2,dotsc,x_{2n-1},x_{2n}) mapsto (x_1+ix_2, dotsc, x_{2n-1}+ix_{2n})$$
    is an isometry, which means that the two spaces are isometric, i.e. they are indistinguishable as metric spaces. Since the usual topologies on both spaces are induced by the metrics, this also means that the two spaces are homeomorphic as topological spaces, which means that they are indistinguishable in that context, as well.



    Therefore any metric or topological property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$. As an example, you have shown that boundedness implies total boundedness in $mathbb{R}^{2n}$, which automatically implies that the same result holds for $mathbb{C}^n$. Similarly, $mathbb{R}^{2n}$ has the Heine-Borel property (a set is compact if and only if it is closed and bounded), which implies that $mathbb{C}^n$ also has this property.



    Do note, however, that we could impose strange metrics or topologies on either space, and we would get something different. For example, the discrete metric on $mathbb{R}$, given by
    $$ d(x,y) = begin{cases} 0 & text{if $x=y$, and} \ 1 & text{otherwise,} end{cases} $$
    turns $mathbb{R}$ into a completely different space.




  • As vector spaces, things get a little more interesting. If we regard both $mathbb{R}^{2n}$ and $mathbb{C}^{n}$ as $mathbb{R}$-vector spaces, then they are indistinguishable. Indeed, there is an "obvious" map sending ${langle 1,0rangle, langle 0,1rangle}$ (a basis for $mathbb{R}^2$) and ${1,i}$ (a basis for $mathbb{C}$ as an $mathbb{R}$-vector space), hence the two $mathbb{R}$-vector spaces are both two dimensional vector spaces over $mathbb{R}$ and therefore isomorphic.



    On the other hand, it is pretty unusual to view $mathbb{C}$ as an $mathbb{R}$-vector space–when we do this, we lose all of the interesting and useful algebraic structure which can be found in the field $mathbb{C}$. Hence it is much more common to view $mathbb{C}^n$ as a $mathbb{C}$-vector space (indeed, if we are working with vector spaces and we aren't doing this, then why would we even bother with the notation $mathbb{C}^n$?). In this setting, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are quite different spaces, owing to the differnet structures of $mathbb{C}$ and $mathbb{R}$. Hence there are things that are true about $mathbb{R}^{2n}$ which are not true about $mathbb{C}^n$, and vice versa. As a simple example, with $n=1$, $mathbb{C}$ is an algebraically closed field, while $mathbb{R}^{2}$ is not even a field.




More generally, your question touches on a branch of mathematics called category theory. I am not very conversant in category theory, so I don't want to say too much about this, but the basic goal of this theory is to study structures in mathematics in an abstract way. If we can show that two objects are isomorphic in some category, then any property which holds for one object will also hold for the other (in that category). Hence if you want to know which properties of $mathbb{R}^{2n}$ easily extend to $mathbb{C}^n$, a good place to start might be to understand the categories in which the two objects are identical, i.e. which structures can be imposed on the two objects which fail to distinguish between them.






share|cite|improve this answer





















  • Thanks alot for this indepth response! I appreciate that!
    – Christian Singer
    Nov 21 at 15:16













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Your basic question could be rephrased in terms of structure. Specifically, "What structures can we impose on $mathbb{R}^{2n}$ and $mathbb{C}^n$ such that the two objects are indistinguishable? What structures would distinguish these two objects?" There are a ton of structures running around here. For example:




  • As sets, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are indistinguishable: there is (at least one) bijective function between the two spaces. In particular, the map $f: mathbb{R}^2 to mathbb{C}$ given by $f(x,y) = x+iy$ is bijective, and we can go from $mathbb{R}^{2n}$ to $mathbb{C}^n$ by applying $f$ componentwise. This implies that the two sets have the same cardinality, which, from the point of view of sets, means that the two objects are the same. Therefore any set theoretic property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$.


  • As topological spaces, $mathbb{R}^{2n}$ and $mathbb{C}^n$. Indeed, this can be understood by seeing that as metric spaces the two objects are the same: in $mathbb{R}^{2n}$ the metric is induced by the norm
    $$ | (x_1, x_2, dotsc x_{2n}) | = left( sum_{j=1}^{2n} x_j^2 right)^{1/2}, $$
    and in $mathbb{C}^n$ the metric is induced by
    $$ | (z_1, z_2, dotsc, z_n) |
    = left( sum_{j=1}^{n} |z_j|^2 right)^{1/2}, $$

    where $|z| = sqrt{Re(z)^2 + Im(z)^2}$ is the modulus of $z$. It is not too difficult to verify that the map
    $$(x_1,x_2,dotsc,x_{2n-1},x_{2n}) mapsto (x_1+ix_2, dotsc, x_{2n-1}+ix_{2n})$$
    is an isometry, which means that the two spaces are isometric, i.e. they are indistinguishable as metric spaces. Since the usual topologies on both spaces are induced by the metrics, this also means that the two spaces are homeomorphic as topological spaces, which means that they are indistinguishable in that context, as well.



    Therefore any metric or topological property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$. As an example, you have shown that boundedness implies total boundedness in $mathbb{R}^{2n}$, which automatically implies that the same result holds for $mathbb{C}^n$. Similarly, $mathbb{R}^{2n}$ has the Heine-Borel property (a set is compact if and only if it is closed and bounded), which implies that $mathbb{C}^n$ also has this property.



    Do note, however, that we could impose strange metrics or topologies on either space, and we would get something different. For example, the discrete metric on $mathbb{R}$, given by
    $$ d(x,y) = begin{cases} 0 & text{if $x=y$, and} \ 1 & text{otherwise,} end{cases} $$
    turns $mathbb{R}$ into a completely different space.




  • As vector spaces, things get a little more interesting. If we regard both $mathbb{R}^{2n}$ and $mathbb{C}^{n}$ as $mathbb{R}$-vector spaces, then they are indistinguishable. Indeed, there is an "obvious" map sending ${langle 1,0rangle, langle 0,1rangle}$ (a basis for $mathbb{R}^2$) and ${1,i}$ (a basis for $mathbb{C}$ as an $mathbb{R}$-vector space), hence the two $mathbb{R}$-vector spaces are both two dimensional vector spaces over $mathbb{R}$ and therefore isomorphic.



    On the other hand, it is pretty unusual to view $mathbb{C}$ as an $mathbb{R}$-vector space–when we do this, we lose all of the interesting and useful algebraic structure which can be found in the field $mathbb{C}$. Hence it is much more common to view $mathbb{C}^n$ as a $mathbb{C}$-vector space (indeed, if we are working with vector spaces and we aren't doing this, then why would we even bother with the notation $mathbb{C}^n$?). In this setting, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are quite different spaces, owing to the differnet structures of $mathbb{C}$ and $mathbb{R}$. Hence there are things that are true about $mathbb{R}^{2n}$ which are not true about $mathbb{C}^n$, and vice versa. As a simple example, with $n=1$, $mathbb{C}$ is an algebraically closed field, while $mathbb{R}^{2}$ is not even a field.




More generally, your question touches on a branch of mathematics called category theory. I am not very conversant in category theory, so I don't want to say too much about this, but the basic goal of this theory is to study structures in mathematics in an abstract way. If we can show that two objects are isomorphic in some category, then any property which holds for one object will also hold for the other (in that category). Hence if you want to know which properties of $mathbb{R}^{2n}$ easily extend to $mathbb{C}^n$, a good place to start might be to understand the categories in which the two objects are identical, i.e. which structures can be imposed on the two objects which fail to distinguish between them.






share|cite|improve this answer












Your basic question could be rephrased in terms of structure. Specifically, "What structures can we impose on $mathbb{R}^{2n}$ and $mathbb{C}^n$ such that the two objects are indistinguishable? What structures would distinguish these two objects?" There are a ton of structures running around here. For example:




  • As sets, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are indistinguishable: there is (at least one) bijective function between the two spaces. In particular, the map $f: mathbb{R}^2 to mathbb{C}$ given by $f(x,y) = x+iy$ is bijective, and we can go from $mathbb{R}^{2n}$ to $mathbb{C}^n$ by applying $f$ componentwise. This implies that the two sets have the same cardinality, which, from the point of view of sets, means that the two objects are the same. Therefore any set theoretic property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$.


  • As topological spaces, $mathbb{R}^{2n}$ and $mathbb{C}^n$. Indeed, this can be understood by seeing that as metric spaces the two objects are the same: in $mathbb{R}^{2n}$ the metric is induced by the norm
    $$ | (x_1, x_2, dotsc x_{2n}) | = left( sum_{j=1}^{2n} x_j^2 right)^{1/2}, $$
    and in $mathbb{C}^n$ the metric is induced by
    $$ | (z_1, z_2, dotsc, z_n) |
    = left( sum_{j=1}^{n} |z_j|^2 right)^{1/2}, $$

    where $|z| = sqrt{Re(z)^2 + Im(z)^2}$ is the modulus of $z$. It is not too difficult to verify that the map
    $$(x_1,x_2,dotsc,x_{2n-1},x_{2n}) mapsto (x_1+ix_2, dotsc, x_{2n-1}+ix_{2n})$$
    is an isometry, which means that the two spaces are isometric, i.e. they are indistinguishable as metric spaces. Since the usual topologies on both spaces are induced by the metrics, this also means that the two spaces are homeomorphic as topological spaces, which means that they are indistinguishable in that context, as well.



    Therefore any metric or topological property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$. As an example, you have shown that boundedness implies total boundedness in $mathbb{R}^{2n}$, which automatically implies that the same result holds for $mathbb{C}^n$. Similarly, $mathbb{R}^{2n}$ has the Heine-Borel property (a set is compact if and only if it is closed and bounded), which implies that $mathbb{C}^n$ also has this property.



    Do note, however, that we could impose strange metrics or topologies on either space, and we would get something different. For example, the discrete metric on $mathbb{R}$, given by
    $$ d(x,y) = begin{cases} 0 & text{if $x=y$, and} \ 1 & text{otherwise,} end{cases} $$
    turns $mathbb{R}$ into a completely different space.




  • As vector spaces, things get a little more interesting. If we regard both $mathbb{R}^{2n}$ and $mathbb{C}^{n}$ as $mathbb{R}$-vector spaces, then they are indistinguishable. Indeed, there is an "obvious" map sending ${langle 1,0rangle, langle 0,1rangle}$ (a basis for $mathbb{R}^2$) and ${1,i}$ (a basis for $mathbb{C}$ as an $mathbb{R}$-vector space), hence the two $mathbb{R}$-vector spaces are both two dimensional vector spaces over $mathbb{R}$ and therefore isomorphic.



    On the other hand, it is pretty unusual to view $mathbb{C}$ as an $mathbb{R}$-vector space–when we do this, we lose all of the interesting and useful algebraic structure which can be found in the field $mathbb{C}$. Hence it is much more common to view $mathbb{C}^n$ as a $mathbb{C}$-vector space (indeed, if we are working with vector spaces and we aren't doing this, then why would we even bother with the notation $mathbb{C}^n$?). In this setting, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are quite different spaces, owing to the differnet structures of $mathbb{C}$ and $mathbb{R}$. Hence there are things that are true about $mathbb{R}^{2n}$ which are not true about $mathbb{C}^n$, and vice versa. As a simple example, with $n=1$, $mathbb{C}$ is an algebraically closed field, while $mathbb{R}^{2}$ is not even a field.




More generally, your question touches on a branch of mathematics called category theory. I am not very conversant in category theory, so I don't want to say too much about this, but the basic goal of this theory is to study structures in mathematics in an abstract way. If we can show that two objects are isomorphic in some category, then any property which holds for one object will also hold for the other (in that category). Hence if you want to know which properties of $mathbb{R}^{2n}$ easily extend to $mathbb{C}^n$, a good place to start might be to understand the categories in which the two objects are identical, i.e. which structures can be imposed on the two objects which fail to distinguish between them.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 at 15:09









Xander Henderson

14.1k103553




14.1k103553












  • Thanks alot for this indepth response! I appreciate that!
    – Christian Singer
    Nov 21 at 15:16


















  • Thanks alot for this indepth response! I appreciate that!
    – Christian Singer
    Nov 21 at 15:16
















Thanks alot for this indepth response! I appreciate that!
– Christian Singer
Nov 21 at 15:16




Thanks alot for this indepth response! I appreciate that!
– Christian Singer
Nov 21 at 15:16


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007735%2fresults-in-mathbbr2n-applying-to-mathbbcn%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Ellipse (mathématiques)

Quarter-circle Tiles

Mont Emei