Prove that for an infinite uncountable set $M$ and infinite countable set $B$: $M sim M cup B$
$begingroup$
I am aware that this has an answer here. But I am wondering whether my proof is also valid.
I am reading the functional analysis book by Kolmogorov and Fomin and they ask to prove the above. But before that, they have shown that:
$$M sim M backslash A_2$$
In the following way:
- Construct a countably infinite set from $M$:
$$A = {a_1, a_2, ..., a_n, ...}$$
- From $A$ construct two other countably infinite sets:
$$A_1 = {a_1, a_3, ..., a_{2n+1}, ... } text{ and } A_2 = {a_2, a_4, ..., a_{2n}, ... }$$
It is trivial to construct a surjective/bijective relationship between $A$ and $A_1$. Now consider that:
$$A cup (M backslash A) = M text{ and } A_1 cup (M backslash A) = M backslash A_2$$
Since $A sim A_1$, we can then also conclude from the above equations that $M sim M backslash A_2$ because we can trivially construct a one-to-one mapping between $(Mbackslash A)$ in the first equation and $(M backslash A)$ in the second equation. Done...
Now from this result, I postulate that:
$$M sim M backslash A_2 iff begin{align*} & M cup A_2 sim (M backslash A_2) cup A_2 \ &M cup A_2 sim M end{align*} tag*{$blacksquare$}$$
Is my proof correct?
EDIT: by the way, I have noticed that in the linked answer they also assume that the countably infinite set is a subspace of the infinite set... I do not really like that answer, there are a couple of things that need clarification there, I doubt I will get an answer there if I comment given how old that question is.
EDIT EDIT (potential proof): I think ignoring all of the above I might come up with a conceptual answer using Hilbert's infinite hotel paradox. First, we take an arbitrary countably infinite set from $M$, let's call it $B$. We now "merge" $B$ and $A$ (some arbitrary countably infinite set that is given). We do this using Hilbert's argument. We move every element/"guest" in set $B$ to it's even room. i.e. $n -> 2n$ (cardinality of naturals and evens is the same) by doing so we create a space for a countably infinite set $A$ inside of $B$. We have now created a countably infinite set $C$ in which every evenly numerated element is from $B$ and every oddly numerated element from $A$. Since the cardinality of $C$ did not change in the process, we have that $M sim M cup A$.
functional-analysis elementary-set-theory
asked Dec 7 '18 at 17:36
i squared - Keep it Reali squared - Keep it Real
1,5851925
$endgroup$
add a comment |
$begingroup$
I am aware that this has an answer here. But I am wondering whether my proof is also valid.
I am reading the functional analysis book by Kolmogorov and Fomin and they ask to prove the above. But before that, they have shown that:
$$M sim M backslash A_2$$
In the following way:
- Construct a countably infinite set from $M$:
$$A = {a_1, a_2, ..., a_n, ...}$$
- From $A$ construct two other countably infinite sets:
$$A_1 = {a_1, a_3, ..., a_{2n+1}, ... } text{ and } A_2 = {a_2, a_4, ..., a_{2n}, ... }$$
It is trivial to construct a surjective/bijective relationship between $A$ and $A_1$. Now consider that:
$$A cup (M backslash A) = M text{ and } A_1 cup (M backslash A) = M backslash A_2$$
Since $A sim A_1$, we can then also conclude from the above equations that $M sim M backslash A_2$ because we can trivially construct a one-to-one mapping between $(Mbackslash A)$ in the first equation and $(M backslash A)$ in the second equation. Done...
Now from this result, I postulate that:
$$M sim M backslash A_2 iff begin{align*} & M cup A_2 sim (M backslash A_2) cup A_2 \ &M cup A_2 sim M end{align*} tag*{$blacksquare$}$$
Is my proof correct?
EDIT: by the way, I have noticed that in the linked answer they also assume that the countably infinite set is a subspace of the infinite set... I do not really like that answer, there are a couple of things that need clarification there, I doubt I will get an answer there if I comment given how old that question is.
EDIT EDIT (potential proof): I think ignoring all of the above I might come up with a conceptual answer using Hilbert's infinite hotel paradox. First, we take an arbitrary countably infinite set from $M$, let's call it $B$. We now "merge" $B$ and $A$ (some arbitrary countably infinite set that is given). We do this using Hilbert's argument. We move every element/"guest" in set $B$ to it's even room. i.e. $n -> 2n$ (cardinality of naturals and evens is the same) by doing so we create a space for a countably infinite set $A$ inside of $B$. We have now created a countably infinite set $C$ in which every evenly numerated element is from $B$ and every oddly numerated element from $A$. Since the cardinality of $C$ did not change in the process, we have that $M sim M cup A$.
functional-analysis elementary-set-theory
asked Dec 7 '18 at 17:36
i squared - Keep it Reali squared - Keep it Real
1,5851925
$endgroup$
$begingroup$
Is $A subset M$ or not?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:10
$begingroup$
Ahh.. good point. Not necessarily.
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:11
$begingroup$
In point 1, you construct $A$ from $M$. Shouldn't $A$ be given (according to the question in the title)?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:14
$begingroup$
Sorry, that $A$ is unrelated..
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:17
add a comment |
$begingroup$
I am aware that this has an answer here. But I am wondering whether my proof is also valid.
I am reading the functional analysis book by Kolmogorov and Fomin and they ask to prove the above. But before that, they have shown that:
$$M sim M backslash A_2$$
In the following way:
- Construct a countably infinite set from $M$:
$$A = {a_1, a_2, ..., a_n, ...}$$
- From $A$ construct two other countably infinite sets:
$$A_1 = {a_1, a_3, ..., a_{2n+1}, ... } text{ and } A_2 = {a_2, a_4, ..., a_{2n}, ... }$$
It is trivial to construct a surjective/bijective relationship between $A$ and $A_1$. Now consider that:
$$A cup (M backslash A) = M text{ and } A_1 cup (M backslash A) = M backslash A_2$$
Since $A sim A_1$, we can then also conclude from the above equations that $M sim M backslash A_2$ because we can trivially construct a one-to-one mapping between $(Mbackslash A)$ in the first equation and $(M backslash A)$ in the second equation. Done...
Now from this result, I postulate that:
$$M sim M backslash A_2 iff begin{align*} & M cup A_2 sim (M backslash A_2) cup A_2 \ &M cup A_2 sim M end{align*} tag*{$blacksquare$}$$
Is my proof correct?
EDIT: by the way, I have noticed that in the linked answer they also assume that the countably infinite set is a subspace of the infinite set... I do not really like that answer, there are a couple of things that need clarification there, I doubt I will get an answer there if I comment given how old that question is.
EDIT EDIT (potential proof): I think ignoring all of the above I might come up with a conceptual answer using Hilbert's infinite hotel paradox. First, we take an arbitrary countably infinite set from $M$, let's call it $B$. We now "merge" $B$ and $A$ (some arbitrary countably infinite set that is given). We do this using Hilbert's argument. We move every element/"guest" in set $B$ to it's even room. i.e. $n -> 2n$ (cardinality of naturals and evens is the same) by doing so we create a space for a countably infinite set $A$ inside of $B$. We have now created a countably infinite set $C$ in which every evenly numerated element is from $B$ and every oddly numerated element from $A$. Since the cardinality of $C$ did not change in the process, we have that $M sim M cup A$.
functional-analysis elementary-set-theory
asked Dec 7 '18 at 17:36
i squared - Keep it Reali squared - Keep it Real
1,5851925
$endgroup$
I am aware that this has an answer here. But I am wondering whether my proof is also valid.
I am reading the functional analysis book by Kolmogorov and Fomin and they ask to prove the above. But before that, they have shown that:
$$M sim M backslash A_2$$
In the following way:
- Construct a countably infinite set from $M$:
$$A = {a_1, a_2, ..., a_n, ...}$$
- From $A$ construct two other countably infinite sets:
$$A_1 = {a_1, a_3, ..., a_{2n+1}, ... } text{ and } A_2 = {a_2, a_4, ..., a_{2n}, ... }$$
It is trivial to construct a surjective/bijective relationship between $A$ and $A_1$. Now consider that:
$$A cup (M backslash A) = M text{ and } A_1 cup (M backslash A) = M backslash A_2$$
Since $A sim A_1$, we can then also conclude from the above equations that $M sim M backslash A_2$ because we can trivially construct a one-to-one mapping between $(Mbackslash A)$ in the first equation and $(M backslash A)$ in the second equation. Done...
Now from this result, I postulate that:
$$M sim M backslash A_2 iff begin{align*} & M cup A_2 sim (M backslash A_2) cup A_2 \ &M cup A_2 sim M end{align*} tag*{$blacksquare$}$$
Is my proof correct?
EDIT: by the way, I have noticed that in the linked answer they also assume that the countably infinite set is a subspace of the infinite set... I do not really like that answer, there are a couple of things that need clarification there, I doubt I will get an answer there if I comment given how old that question is.
EDIT EDIT (potential proof): I think ignoring all of the above I might come up with a conceptual answer using Hilbert's infinite hotel paradox. First, we take an arbitrary countably infinite set from $M$, let's call it $B$. We now "merge" $B$ and $A$ (some arbitrary countably infinite set that is given). We do this using Hilbert's argument. We move every element/"guest" in set $B$ to it's even room. i.e. $n -> 2n$ (cardinality of naturals and evens is the same) by doing so we create a space for a countably infinite set $A$ inside of $B$. We have now created a countably infinite set $C$ in which every evenly numerated element is from $B$ and every oddly numerated element from $A$. Since the cardinality of $C$ did not change in the process, we have that $M sim M cup A$.
functional-analysis elementary-set-theory
functional-analysis elementary-set-theory
asked Dec 7 '18 at 17:36
i squared - Keep it Reali squared - Keep it Real
1,5851925
asked Dec 7 '18 at 17:36
i squared - Keep it Reali squared - Keep it Real
1,5851925
edited Dec 9 '18 at 23:20
i squared - Keep it Real
asked Dec 7 '18 at 17:36
i squared - Keep it Reali squared - Keep it Real
1,5851925
asked Dec 7 '18 at 17:36
i squared - Keep it Reali squared - Keep it Real
1,5851925
asked Dec 7 '18 at 17:36
i squared - Keep it Reali squared - Keep it Real
1,5851925
1,5851925
$begingroup$
Is $A subset M$ or not?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:10
$begingroup$
Ahh.. good point. Not necessarily.
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:11
$begingroup$
In point 1, you construct $A$ from $M$. Shouldn't $A$ be given (according to the question in the title)?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:14
$begingroup$
Sorry, that $A$ is unrelated..
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:17
add a comment |
$begingroup$
Is $A subset M$ or not?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:10
$begingroup$
Ahh.. good point. Not necessarily.
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:11
$begingroup$
In point 1, you construct $A$ from $M$. Shouldn't $A$ be given (according to the question in the title)?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:14
$begingroup$
Sorry, that $A$ is unrelated..
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:17
$begingroup$
Is $A subset M$ or not?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:10
$begingroup$
Is $A subset M$ or not?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:10
$begingroup$
Ahh.. good point. Not necessarily.
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:11
$begingroup$
Ahh.. good point. Not necessarily.
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:11
$begingroup$
In point 1, you construct $A$ from $M$. Shouldn't $A$ be given (according to the question in the title)?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:14
$begingroup$
In point 1, you construct $A$ from $M$. Shouldn't $A$ be given (according to the question in the title)?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:14
$begingroup$
Sorry, that $A$ is unrelated..
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:17
$begingroup$
Sorry, that $A$ is unrelated..
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:17
add a comment |
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$begingroup$
Is $A subset M$ or not?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:10
$begingroup$
Ahh.. good point. Not necessarily.
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:11
$begingroup$
In point 1, you construct $A$ from $M$. Shouldn't $A$ be given (according to the question in the title)?
$endgroup$
– Alex Vong
Dec 7 '18 at 18:14
$begingroup$
Sorry, that $A$ is unrelated..
$endgroup$
– i squared - Keep it Real
Dec 7 '18 at 18:17