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I was reading this question which shows that for metric spaces, $$M \text{ compact} \iff \big((\text{Graph}(\varphi) \text{ closed} \implies \varphi \text{ continuous}) \,\, \forall \text{ set maps }\varphi:N \to M \big),$$ where $N$ is assumed to be a metric space as well.

I was trying to make it work for non-metric spaces. ($\implies$) works fine. In showing ($\impliedby$), I tried to work by contrapositive and let $m_\alpha$ be a net with no convergent subnet, give the directed set $A$ the ordered topology, add a final point $\beta$ to $A$, and use the net $\Delta: A \cup \{\beta\} \to M$ where $\Delta(\beta)$ is some arbitrary point in $M$. I'd like this to have closed graph but to be discontinuous.

I believe it is clearly discontinuous, but I'm not sure the graph is closed. Does anyone have a tip, or a counterexample to show that ($\impliedby$) does not hold?

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Eric Auld
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1 Answers1

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I shall follow Exercise 3.12.5 from [Engelking1989].

A Hausdorff space $X$ is called H-minimal if every one-to-one continuous map of $X$ onto a Hausdorff space is a homeomorphism. I can show the following

Proposition. Let $X$ be a Hausdorff space such that for any Hausdorff space $Y$ any map $f:Y\to X$ with closed graph is continuous. Then the space $X$ is H-minimal.

Proof. Assume the converse. Let $X$ be a Hausdorff space which is not H-mininmal. Then there exists a Hausdorff space $Y$ and one-to-one continuous map $g: X\to Y$ onto which is not a homeomorphism. It is easy to check that a map $g^{-1}:Y\to X$ has a closed graph, but it is not continuous, a contradiction. $\square$

A Hausdorff space $X$ is H-minimal iff $X$ is semiregular and H-closed, and there exists a non-compact H-minimal space. I recall that a space $X$ is called semiregular if $X$ is Hausdorff and the family of all open domains is a base for $X$. A subset $U$ of a space is called an open domain provided $U=\operatorname{int}\overline{U}$. Every regular space is semiregular, but there exists a Hausdorff space that is not semiregular and a semiregular space that is not regular, also there exist $T_1$-spaces in which open domains form a base but which are not Hausdorff [Engelking1989, Ex. 1.7.8]. It is well known that a compact Hausdorff space $X$ is a closed subspace of any Hausdorff space which contains $X$. So a Hausdorff space $X$ is defined to be H-closed provided $X$ is a closed subspace of any Hausdorff space which contains $X$. A regular space is H-closed if and only if it is compact, but there exists a non-regular H-closed space.

Theorem. ([AlexandroffUrysohn1929], (announcement in [AlexandroffUrysohn1923] and [Aleksandrov1939])). For a Hausdorff space $X$ the following conditions are equivalent:

  1. The space $X$ is H-closed;

  2. For every family $\left\{V_s\colon s\in S\right\}$ of open subsets of $X$ which has the finite intersection property the intersection $\bigcap\left\{\overline{V_s}\colon s\in S\right\}$ is non-empty;

  3. Every ultrafilter in the family of all open subsets of $X$ converges;

  4. Every open cover $\left\{U_s\colon s\in S\right\}$ of the space $X$ contains a finite subfamily $\left\{U_{s_1},U_{s_2},\ldots, U_{s_n}\right\}$ such that $\overline{U_{s_1}}\cup \overline{U_{s_2}}\cup\dots\cup\overline{U_{s_n}}=X$.$\square$

References

[AlexandroffUrysohn1923] P. Alexandroff and P. Urysohn, Sur les espaces topologiques compacts, Bull. Intern. Acad. Pol. Sci. Sér. A (1923), 5-8.

[AlexandroffUrysohn1929] P. Alexandroff and P. Urysohn, Mémoire sur les espaces topologiques compacts, Verh. Akad. Wetensch. Amsterdam 14 (1929).

[Aleksandrov1939] P. S. Aleksandrov, On bicompact extensions of topological spaces, Mat. Sb. (N.S.) 5 (47) (1939), 403--423. (in Russian)

[Engelking1989] R. Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

Alex Ravsky
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