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Can you give me an example of a non compact topological space with compact dense subset?

I know a Hausdorff topological space $(X, \tau)$ with compact dense subset must be compact.

Hence, my intuition suggests that there may be a non compact topological space with compact dense subset.

And such space will not be Hausdorff space anymore.

But it is difficult for me to cite this example.

Please explain it in details. Thanks.

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    $\begingroup$ Take $\mathbb{N}$ with the open sets $\{0\}\cup A$ for $A\subset\mathbb{N}$. It is non-compact, since $\{\{0,x\}:\ x\in\mathbb{N}\}$ doesn't have a finite subset that is a cover. The set $\{0\}$ is dense and compact. $\endgroup$
    – plop
    Nov 24, 2021 at 11:59

2 Answers 2

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Take the included point topology on e..g. $X=\Bbb R$: all open subsets are $\emptyset$ and all $A \subseteq \Bbb R$ with $0 \in A$.

Then $\{0\}$ is compact and dense while $X$ is not compact: consider the open cover $\{\{0,x\}, x \in X\}$.

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    $\begingroup$ Any infinite set $X$ will do of course. $\endgroup$ Nov 24, 2021 at 12:02
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You can construct a nice example by modifying the "real line with two $0$'s" as follows:

Let $Y=[-1,1]\times \mathbb{N}$ and take $X$ to be the quotient space $Y/\mathord{\sim}$ where we identify $(t,n)\sim (t,m)$ for all $t\neq 0$ and $m,n\in\mathbb{N}$. Essentially, you collapse all of the intervals together except at $0$ you leave the points alone. The space $X$ is a single interval but where you have a countably infinite number of $0$'s. Let's see why $X$ is a $T_1$, non-compact space which has a compact, open subset.

Let $q:Y\to X$ denote the quotient map. It's too not hard to check that $q$ is an open map. Let's let $0_n=q(0,n)$ denote the "$n$-th zero" and identify $X\backslash \{0_n\mid n\in\mathbb{N}\}$ with $[-1,0)\cup (0,1]$ since $q$ maps $[-1,0)\cup (0,1]\times \{n\}$ homeomorphically onto its image.

$X$ is $T_1$ since the fibers of $q$ are of the form $q^{-1}(t)=\{t\}\times \mathbb{N}$, $t\neq 0$ or $q^{-1}(0_n)=\{(0,n)\}$, which are closed in $[-1,1]\times\mathbb{N}$.

Since $q$ is an open map, each interval $U_n=q([-1,1]\times\{n\})$ is open neighborhood of $0_n$ in $X$. Moreover, $U_n$ is dense in $X$ because every open neighborhood of $x_m$ will meet $U_n$. Finally, $U_n$ is compact because it's the continuous image of the compact set $[-1,1]\times\{n\}$.

The collection $\{U_n\mid n\in\mathbb{N}\}$ is an open cover of $X$, which has no finite subcover and so $X$ is not compact.

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  • $\begingroup$ $q((\frac 1n,0)_n$ converges to all $0_k$ simultaneously right, so $X$ is not US? $\endgroup$ Jan 16, 2022 at 22:55

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