Locally compact in $mathbb{R}^n$ and Hausdorff spaces
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I have to show that
(1) Show that $mathbb{R}^n$ is locally compact.
https://i.imgur.com/GL4PJj5.png
I have provided a link with my answer since it's long and would take me forever to write in here.
Is this proof good enough or do I also need to show that $mathbb{R}^n$$^-$$^1$ and $mathbb{R}$ is homeomorphic to $mathbb{R}^n$?
And as a side question: is there a shorter/easier proof of this?
(2) Assume that X and Y are locally compact Hausdorff spaces. Show that X×Y is locally compact.
Since I proved in (1) that X×Y is locally compact if X and Y are locally compact, do I just refer to that here?
Thank you for your help!
Also: I have provided our definition of being locally compact in my link to (1).
general-topology geometry geometric-topology
asked Nov 22 at 15:31
Nikolaj
436
add a comment |
up vote
-1
down vote
favorite
I have to show that
(1) Show that $mathbb{R}^n$ is locally compact.
https://i.imgur.com/GL4PJj5.png
I have provided a link with my answer since it's long and would take me forever to write in here.
Is this proof good enough or do I also need to show that $mathbb{R}^n$$^-$$^1$ and $mathbb{R}$ is homeomorphic to $mathbb{R}^n$?
And as a side question: is there a shorter/easier proof of this?
(2) Assume that X and Y are locally compact Hausdorff spaces. Show that X×Y is locally compact.
Since I proved in (1) that X×Y is locally compact if X and Y are locally compact, do I just refer to that here?
Thank you for your help!
Also: I have provided our definition of being locally compact in my link to (1).
general-topology geometry geometric-topology
asked Nov 22 at 15:31
Nikolaj
436
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have to show that
(1) Show that $mathbb{R}^n$ is locally compact.
https://i.imgur.com/GL4PJj5.png
I have provided a link with my answer since it's long and would take me forever to write in here.
Is this proof good enough or do I also need to show that $mathbb{R}^n$$^-$$^1$ and $mathbb{R}$ is homeomorphic to $mathbb{R}^n$?
And as a side question: is there a shorter/easier proof of this?
(2) Assume that X and Y are locally compact Hausdorff spaces. Show that X×Y is locally compact.
Since I proved in (1) that X×Y is locally compact if X and Y are locally compact, do I just refer to that here?
Thank you for your help!
Also: I have provided our definition of being locally compact in my link to (1).
general-topology geometry geometric-topology
asked Nov 22 at 15:31
Nikolaj
436
I have to show that
(1) Show that $mathbb{R}^n$ is locally compact.
https://i.imgur.com/GL4PJj5.png
I have provided a link with my answer since it's long and would take me forever to write in here.
Is this proof good enough or do I also need to show that $mathbb{R}^n$$^-$$^1$ and $mathbb{R}$ is homeomorphic to $mathbb{R}^n$?
And as a side question: is there a shorter/easier proof of this?
(2) Assume that X and Y are locally compact Hausdorff spaces. Show that X×Y is locally compact.
Since I proved in (1) that X×Y is locally compact if X and Y are locally compact, do I just refer to that here?
Thank you for your help!
Also: I have provided our definition of being locally compact in my link to (1).
general-topology geometry geometric-topology
general-topology geometry geometric-topology
asked Nov 22 at 15:31
Nikolaj
436
asked Nov 22 at 15:31
Nikolaj
436
asked Nov 22 at 15:31
Nikolaj
436
asked Nov 22 at 15:31
Nikolaj
436
asked Nov 22 at 15:31
Nikolaj
436
436
add a comment |
add a comment |
1 Answer
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In $mathbb{R}^n$, by Heine-Borel a set is compact iff it is closed and bounded.
So any open ball $B(x,varepsilon) ={y: d(x,y) < varepsilon}$ neighbourhood of $x$ contains a compact neighbourhood of $x$, namely the closed ball $D(x,delta) = {y: d(x,y) le delta}$ for $delta < varepsilon$ (bounded by definition, closed by a theorem).
(2) is a consequence of the fact that a product of two compact neighbourhoods is a compact neighbourhood in the product.
answered Nov 22 at 16:58
Henno Brandsma
103k345112
Oh, so to (1) can you just argue that since any open ball $B(x,ε)$ in $mathbb{R}^n$ has a compact closure $bar{B}$$(x,ε)$$={{yinmathbb{R}^n:d(x,y)<ε}}$ then $mathbb{R}^n$ is locally compact?
– Nikolaj
Nov 22 at 17:56
@Nikolaj yes, that’s valid too. It depends on your definition of local compactness.
– Henno Brandsma
Nov 22 at 18:05
Our definition is that X is called locally compact if there for any p in X exists a neighbourhood U of p such that the closure $bar{U}$ is compact and a subset of X
– Nikolaj
Nov 22 at 18:13
@Nikolaj in some definitions you have to show there is a local base of compact neighbourhoods.
– Henno Brandsma
Nov 22 at 18:14
add a comment |
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In $mathbb{R}^n$, by Heine-Borel a set is compact iff it is closed and bounded.
So any open ball $B(x,varepsilon) ={y: d(x,y) < varepsilon}$ neighbourhood of $x$ contains a compact neighbourhood of $x$, namely the closed ball $D(x,delta) = {y: d(x,y) le delta}$ for $delta < varepsilon$ (bounded by definition, closed by a theorem).
(2) is a consequence of the fact that a product of two compact neighbourhoods is a compact neighbourhood in the product.
answered Nov 22 at 16:58
Henno Brandsma
103k345112
Oh, so to (1) can you just argue that since any open ball $B(x,ε)$ in $mathbb{R}^n$ has a compact closure $bar{B}$$(x,ε)$$={{yinmathbb{R}^n:d(x,y)<ε}}$ then $mathbb{R}^n$ is locally compact?
– Nikolaj
Nov 22 at 17:56
@Nikolaj yes, that’s valid too. It depends on your definition of local compactness.
– Henno Brandsma
Nov 22 at 18:05
Our definition is that X is called locally compact if there for any p in X exists a neighbourhood U of p such that the closure $bar{U}$ is compact and a subset of X
– Nikolaj
Nov 22 at 18:13
@Nikolaj in some definitions you have to show there is a local base of compact neighbourhoods.
– Henno Brandsma
Nov 22 at 18:14
add a comment |
up vote
1
down vote
In $mathbb{R}^n$, by Heine-Borel a set is compact iff it is closed and bounded.
So any open ball $B(x,varepsilon) ={y: d(x,y) < varepsilon}$ neighbourhood of $x$ contains a compact neighbourhood of $x$, namely the closed ball $D(x,delta) = {y: d(x,y) le delta}$ for $delta < varepsilon$ (bounded by definition, closed by a theorem).
(2) is a consequence of the fact that a product of two compact neighbourhoods is a compact neighbourhood in the product.
answered Nov 22 at 16:58
Henno Brandsma
103k345112
Oh, so to (1) can you just argue that since any open ball $B(x,ε)$ in $mathbb{R}^n$ has a compact closure $bar{B}$$(x,ε)$$={{yinmathbb{R}^n:d(x,y)<ε}}$ then $mathbb{R}^n$ is locally compact?
– Nikolaj
Nov 22 at 17:56
@Nikolaj yes, that’s valid too. It depends on your definition of local compactness.
– Henno Brandsma
Nov 22 at 18:05
Our definition is that X is called locally compact if there for any p in X exists a neighbourhood U of p such that the closure $bar{U}$ is compact and a subset of X
– Nikolaj
Nov 22 at 18:13
@Nikolaj in some definitions you have to show there is a local base of compact neighbourhoods.
– Henno Brandsma
Nov 22 at 18:14
add a comment |
up vote
1
down vote
up vote
1
down vote
In $mathbb{R}^n$, by Heine-Borel a set is compact iff it is closed and bounded.
So any open ball $B(x,varepsilon) ={y: d(x,y) < varepsilon}$ neighbourhood of $x$ contains a compact neighbourhood of $x$, namely the closed ball $D(x,delta) = {y: d(x,y) le delta}$ for $delta < varepsilon$ (bounded by definition, closed by a theorem).
(2) is a consequence of the fact that a product of two compact neighbourhoods is a compact neighbourhood in the product.
answered Nov 22 at 16:58
Henno Brandsma
103k345112
In $mathbb{R}^n$, by Heine-Borel a set is compact iff it is closed and bounded.
So any open ball $B(x,varepsilon) ={y: d(x,y) < varepsilon}$ neighbourhood of $x$ contains a compact neighbourhood of $x$, namely the closed ball $D(x,delta) = {y: d(x,y) le delta}$ for $delta < varepsilon$ (bounded by definition, closed by a theorem).
(2) is a consequence of the fact that a product of two compact neighbourhoods is a compact neighbourhood in the product.
answered Nov 22 at 16:58
Henno Brandsma
103k345112
answered Nov 22 at 16:58
Henno Brandsma
103k345112
answered Nov 22 at 16:58
Henno Brandsma
103k345112
answered Nov 22 at 16:58
Henno Brandsma
103k345112
103k345112
Oh, so to (1) can you just argue that since any open ball $B(x,ε)$ in $mathbb{R}^n$ has a compact closure $bar{B}$$(x,ε)$$={{yinmathbb{R}^n:d(x,y)<ε}}$ then $mathbb{R}^n$ is locally compact?
– Nikolaj
Nov 22 at 17:56
@Nikolaj yes, that’s valid too. It depends on your definition of local compactness.
– Henno Brandsma
Nov 22 at 18:05
Our definition is that X is called locally compact if there for any p in X exists a neighbourhood U of p such that the closure $bar{U}$ is compact and a subset of X
– Nikolaj
Nov 22 at 18:13
@Nikolaj in some definitions you have to show there is a local base of compact neighbourhoods.
– Henno Brandsma
Nov 22 at 18:14
add a comment |
Oh, so to (1) can you just argue that since any open ball $B(x,ε)$ in $mathbb{R}^n$ has a compact closure $bar{B}$$(x,ε)$$={{yinmathbb{R}^n:d(x,y)<ε}}$ then $mathbb{R}^n$ is locally compact?
– Nikolaj
Nov 22 at 17:56
@Nikolaj yes, that’s valid too. It depends on your definition of local compactness.
– Henno Brandsma
Nov 22 at 18:05
Our definition is that X is called locally compact if there for any p in X exists a neighbourhood U of p such that the closure $bar{U}$ is compact and a subset of X
– Nikolaj
Nov 22 at 18:13
@Nikolaj in some definitions you have to show there is a local base of compact neighbourhoods.
– Henno Brandsma
Nov 22 at 18:14
Oh, so to (1) can you just argue that since any open ball $B(x,ε)$ in $mathbb{R}^n$ has a compact closure $bar{B}$$(x,ε)$$={{yinmathbb{R}^n:d(x,y)<ε}}$ then $mathbb{R}^n$ is locally compact?
– Nikolaj
Nov 22 at 17:56
Oh, so to (1) can you just argue that since any open ball $B(x,ε)$ in $mathbb{R}^n$ has a compact closure $bar{B}$$(x,ε)$$={{yinmathbb{R}^n:d(x,y)<ε}}$ then $mathbb{R}^n$ is locally compact?
– Nikolaj
Nov 22 at 17:56
@Nikolaj yes, that’s valid too. It depends on your definition of local compactness.
– Henno Brandsma
Nov 22 at 18:05
@Nikolaj yes, that’s valid too. It depends on your definition of local compactness.
– Henno Brandsma
Nov 22 at 18:05
Our definition is that X is called locally compact if there for any p in X exists a neighbourhood U of p such that the closure $bar{U}$ is compact and a subset of X
– Nikolaj
Nov 22 at 18:13
Our definition is that X is called locally compact if there for any p in X exists a neighbourhood U of p such that the closure $bar{U}$ is compact and a subset of X
– Nikolaj
Nov 22 at 18:13
@Nikolaj in some definitions you have to show there is a local base of compact neighbourhoods.
– Henno Brandsma
Nov 22 at 18:14
@Nikolaj in some definitions you have to show there is a local base of compact neighbourhoods.
– Henno Brandsma
Nov 22 at 18:14
add a comment |
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