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If $X$ a topological space one says that $X$ is universally closed if for every Hausdorff space $Y$ and every (continuous) map $f:X\rightarrow Y$, the image of $X$ is a closed subset of $Y$.

It is clear that every compact space is universally closed, but are there non compact universally closed spaces?

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H-closed spaces are defined as those Hausdorff spaces $X$ such that for every embedding $e$ from $X$ into a Hausdorff space $Y$ has a closed image. It is well-known that there are H-closed spaces that are not compact (a characterisation: every open cover of $X$ has a finite subcover whose union is dense in $X$, which shows the relation to compactness even better). By the latter characterisation, we see that a regular H-closed space is compact. Now, I wonder, is every H-closed space universally closed, i.e. do these classes coincide for Hausdorff spaces? – Henno Brandsma Oct 21 '10 at 17:39
up vote 12 down vote accepted

If $Z$ is not compact, and $X=\{p\}\cup Z$ is the space whose nonempty open sets are of the form $\{p\}\cup V$ with $V$ open in $Z$, then $X$ is not compact, but every continuous function from $X$ to a Hausdorff space is constant.

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Taking $V$ empty, you get that $X=Z\coprod\{p\}$! – Laurent Moret-Bailly Oct 21 '10 at 10:40
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Laurent, he hasn't included the old open sets, but just added $p$ to them all, so what you say is not correct. If $Z$ is nonempty, for example, then $X$ is not Hausdorff, even if $Z$ is. – Joel David Hamkins Oct 21 '10 at 10:59
    
Sorry, I see the point. – Laurent Moret-Bailly Oct 21 '10 at 16:29
    
As a generalization, any hyperconnected (for general topologist) / irreducible (for algebraic geometer) space is universally closed. – Fan Zheng May 3 at 22:52

No: if $X$ is non compact, it is a proper and dense subset, thus not closed, in its Stone–Čech compactification.

[edit] This is ok e.g. if X is $T_{3.5}$, as Bruno observes, otherwise $X\to\beta X$ may be surjective (I tend to culpably remove from my conscience the existence of less-separated topological spaces). So, given that a non-compact $T_{3.5}$ space is a proper subspace of its SC compactification, a suitable statement is "compact equals universally closed for $T_{3.5}$ spaces ", and analogous statements may be sought for other categories of topological spaces. In any case, it does not seem very fair, in the definition of "universally closed", to ask for more separation in $Y$ than in $X$ as you are doing. As in Jonas Meyer's answer: for instance, let's consider $X= \mathbb{Z} ,$ with left-unbounded order intervals as open sets. Then every two non-empty closed sets have non-empty intersection, so any continuous map $f:X\to Y$ to a Hausdorff space $Y$ is constant, thus $X$ is $T_0$, non-compact but universally closed in the definition you gave).

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Note: If X is compact, it is also dense in its Stone-Cech compactification... but the latter is simply equal to X. – André Henriques Oct 20 '10 at 23:56
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You need to make assumptions on $X$ in order to view it as a subspace of its Stone-Čech compactification. But for general spaces I think your observation works by considering the image of $X$ in $\beta X$. – Joel David Hamkins Oct 21 '10 at 0:06
    
Yes, of course, I mean $f$ to be the compactification $X\to\beta X$, (that is not injective in general). – Pietro Majer Oct 21 '10 at 5:32
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If $X$ is Tychonoff then $f$ is not surjective and everything is ok. However, in general the map $f$ might be surjective: this holds in the example below from Meyer, where the Stone–Čech compactification of $X$ is a point. – Bruno Martelli Oct 21 '10 at 9:02
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Let me correct my comment: I used the fact that every point has a basis of closed neighborhoods, which means $X$ is regular, not just Hausdorff. See the coment by Henno Brandsma. – Laurent Moret-Bailly Oct 23 '10 at 15:50

I believe the question whether $X$ can be chosen to be Hausdorff was left open by both existing answers. The solution is provided by the H-closed spaces of Henno Brandsma's comment. I shall answer the question from that comment in the positive.

Proposition. A Hausdorff space $X$ is H-closed iff it is universally closed.

Proof. The "if" direction is trivial.

To prove "only if", we use the well-known fact that a Hausdorff space is H-closed iff every open cover of $X$ contains a finite collection of sets whose closures cover $X$. (See the Wikipedia page cited above, or the section on H-closed spaces in Extensions and Absolutes of Hausdorff Spaces by Jack R. Porter and R. Grant Woods.)

So suppose $f\colon X\to Y$ is continuous, where $Y$ is Hausdorff, and let $y\in Y\setminus f(X)$. Since $Y$ is Hausdorff, for every $x\in X$ there is a closed neighbourhood $U[x]\subset Y$ of $f(x)$ not containing $y$. Since $X$ is H-closed, there is a finite collection $x_1,\dots,x_k$ such that $$ X = \bigcup_i f^{-1}(U[x_i]). $$ So $$ y\in Y\setminus \bigcup_i U[x_i] \subset Y\setminus f(X), $$ and hence $Y\setminus f(X)$ is a neighbourhood of $y$. So $f(X)$ is closed, as claimed. $\blacksquare$

Alternatively, you may consider directly the following simple example of a H-closed space from Porter-Woods. This space is given by $$ X := \{p^-,p^+\} \cup \{(1/n,1/m)\colon n\in\mathbb{N}, m\in\mathbb{Z}\setminus\{0\}\} \cup \{(1/n,0)\colon n\in\mathbb{N}\}.$$ $Y:=X\setminus\{p^-,p^+\}$ has the usual topology induced from $\mathbb{R}^2$, while a neighbourhood of $p^+$ (resp. $p^-$) should contain all points $(1/n,1/m)$ with sufficiently large $n$ and positive (resp. negative) $m$.

It is easy to verify directly that this space is universally closed (and H-closed).

As noted in Pietro's answer, the map from a universally closed space to its Stone-Cech compactification is surjective. The question remains whether there is a space with the latter property which is not H-closed.

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