Proving every compact space is locally compact.
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A topological space is said to be locally compact if each point $xin X$ has at least one neighbourhood which is compact.
Prove every compact space is locally compact.
I thought this problem was trivial but I am not sure. If I consider $(X,tau)$ to be compact topological space then every $xin X$ has X as neighbourhood which implies $(X,tau)$ is locally compact.
Question:
Is this reasoning right?
Thanks in advance!
general-topology proof-verification compactness
asked Dec 24 '18 at 19:46
Pedro GomesPedro Gomes
2,0162823
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add a comment |
$begingroup$
A topological space is said to be locally compact if each point $xin X$ has at least one neighbourhood which is compact.
Prove every compact space is locally compact.
I thought this problem was trivial but I am not sure. If I consider $(X,tau)$ to be compact topological space then every $xin X$ has X as neighbourhood which implies $(X,tau)$ is locally compact.
Question:
Is this reasoning right?
Thanks in advance!
general-topology proof-verification compactness
asked Dec 24 '18 at 19:46
Pedro GomesPedro Gomes
2,0162823
$endgroup$
6
$begingroup$
Yes, it's trivial (as the terminology would indicate). Your reasoning is correct.
$endgroup$
– zoidberg
Dec 24 '18 at 19:50
$begingroup$
As stated it's trivial because $X$ is a nbhd of each $xin X$. But this is not the usual definition of locally compact. Caution: There is more than one def'n, although they are equivalent for Hausdorff spaces.
$endgroup$
– DanielWainfleet
Dec 24 '18 at 23:08
add a comment |
$begingroup$
A topological space is said to be locally compact if each point $xin X$ has at least one neighbourhood which is compact.
Prove every compact space is locally compact.
I thought this problem was trivial but I am not sure. If I consider $(X,tau)$ to be compact topological space then every $xin X$ has X as neighbourhood which implies $(X,tau)$ is locally compact.
Question:
Is this reasoning right?
Thanks in advance!
general-topology proof-verification compactness
asked Dec 24 '18 at 19:46
Pedro GomesPedro Gomes
2,0162823
$endgroup$
A topological space is said to be locally compact if each point $xin X$ has at least one neighbourhood which is compact.
Prove every compact space is locally compact.
I thought this problem was trivial but I am not sure. If I consider $(X,tau)$ to be compact topological space then every $xin X$ has X as neighbourhood which implies $(X,tau)$ is locally compact.
Question:
Is this reasoning right?
Thanks in advance!
general-topology proof-verification compactness
general-topology proof-verification compactness
asked Dec 24 '18 at 19:46
Pedro GomesPedro Gomes
2,0162823
asked Dec 24 '18 at 19:46
Pedro GomesPedro Gomes
2,0162823
asked Dec 24 '18 at 19:46
Pedro GomesPedro Gomes
2,0162823
asked Dec 24 '18 at 19:46
Pedro GomesPedro Gomes
2,0162823
asked Dec 24 '18 at 19:46
Pedro GomesPedro Gomes
2,0162823
2,0162823
6
$begingroup$
Yes, it's trivial (as the terminology would indicate). Your reasoning is correct.
$endgroup$
– zoidberg
Dec 24 '18 at 19:50
$begingroup$
As stated it's trivial because $X$ is a nbhd of each $xin X$. But this is not the usual definition of locally compact. Caution: There is more than one def'n, although they are equivalent for Hausdorff spaces.
$endgroup$
– DanielWainfleet
Dec 24 '18 at 23:08
add a comment |
6
$begingroup$
Yes, it's trivial (as the terminology would indicate). Your reasoning is correct.
$endgroup$
– zoidberg
Dec 24 '18 at 19:50
$begingroup$
As stated it's trivial because $X$ is a nbhd of each $xin X$. But this is not the usual definition of locally compact. Caution: There is more than one def'n, although they are equivalent for Hausdorff spaces.
$endgroup$
– DanielWainfleet
Dec 24 '18 at 23:08
6
6
$begingroup$
Yes, it's trivial (as the terminology would indicate). Your reasoning is correct.
$endgroup$
– zoidberg
Dec 24 '18 at 19:50
$begingroup$
Yes, it's trivial (as the terminology would indicate). Your reasoning is correct.
$endgroup$
– zoidberg
Dec 24 '18 at 19:50
$begingroup$
As stated it's trivial because $X$ is a nbhd of each $xin X$. But this is not the usual definition of locally compact. Caution: There is more than one def'n, although they are equivalent for Hausdorff spaces.
$endgroup$
– DanielWainfleet
Dec 24 '18 at 23:08
$begingroup$
As stated it's trivial because $X$ is a nbhd of each $xin X$. But this is not the usual definition of locally compact. Caution: There is more than one def'n, although they are equivalent for Hausdorff spaces.
$endgroup$
– DanielWainfleet
Dec 24 '18 at 23:08
add a comment |
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6
$begingroup$
Yes, it's trivial (as the terminology would indicate). Your reasoning is correct.
$endgroup$
– zoidberg
Dec 24 '18 at 19:50
$begingroup$
As stated it's trivial because $X$ is a nbhd of each $xin X$. But this is not the usual definition of locally compact. Caution: There is more than one def'n, although they are equivalent for Hausdorff spaces.
$endgroup$
– DanielWainfleet
Dec 24 '18 at 23:08