Locally compact in $mathbb{R}^n$ and Hausdorff spaces
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
386
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
386
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
386
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
386
asked Nov 22 at 15:31
Nikolaj
386
asked Nov 22 at 15:31
Nikolaj
386
asked Nov 22 at 15:31
Nikolaj
386
asked Nov 22 at 15:31
Nikolaj
386
386
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
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
102k344108
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 |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
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
102k344108
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
0
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
102k344108
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
0
down vote
up vote
0
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
102k344108
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
102k344108
answered Nov 22 at 16:58
Henno Brandsma
102k344108
answered Nov 22 at 16:58
Henno Brandsma
102k344108
answered Nov 22 at 16:58
Henno Brandsma
102k344108
102k344108
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 |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009269%2flocally-compact-in-mathbbrn-and-hausdorff-spaces%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
