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I have difficulties to answer at that question:

Let $X$ be a Hausdorff and compact topological space, and let $Y$ be a topological space. Let $f:X→Y$ be such that $G(f) = \{(x,f(x))|x∈X\} $ is a compact subset of $X \times Y$. Show that f is continuous.

I tried using the restrictions of the projections to $G(f)$, but I am block. I don't know how to use the Hausdorff condition in such a problem!

eti902
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3 Answers3

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You are on the right track. What you want is that the projection map from the graph to $X$ is a homeomorphism. The relevant theorem is that a continuous bijection from a (quasi-)compact space to a Hausdorff space is a homeomorphism.

Carsten S
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  • This general approach works, in the sense that the projection map restricts to a homeomorphism from the graph to $X$, but the theorem that a continuous bijection from a quasi-compact space to a Hausdorff space doesn't seem directly useful, because neither of $X$ or $Y$ is required to be Hausdorff. (If $Y$ is Hausdorff then the continuity of $f$ implies its graph is closed, but here we start with the hypothesis that the graph is closed.) – krm2233 Jul 24 '23 at 15:28
  • @krm2233, the question states that $X$ is hausdorff. The theorem is applied to $G(f)\to X$. – Carsten S Jul 24 '23 at 15:37
  • Ah sorry, I see I misread the question, or rather I read the title of the post and assumed I knew what the question was, when in fact the title has nothing to do with the question: It is not a closed graph criterion for continuity, it is a "compact-Hausdorff" graph criterion: $G(f)$ is not closed in general: e.g. if $Y = {0,1}$ with topology ${\emptyset,{0},{0,1}}$, then if $X = {}$ is a 1-point set and $f$ is given by $f()=0$, then $X$ and $G(f)$ compact and Hausdorff (both are 1-point sets) but $G(f)\subset X\times Y$ has $\overline{G(f)} = X\times Y$, so $G(f)$ is not closed. – krm2233 Jul 24 '23 at 18:40
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To show $f$ is continuous, let $C$ be a closed subset of $Y$. Suppose $\{ x_n\} \in f^{-1}(C)$ and $x_n \to x$. Then $(x_n,f(x_n)) \in G(f)$, and this has a convergent subsequence $(x_{n_k},f({x_{n_k}}))$ with limit $(x,y) \in G(f)$, so $y=f(x) \in X$ and since $f({x_{n_k}})) \to y$, it follows that $y=f(x) \in C$. Thus $x \in f^{-1}(C)$. Therefore $f^{-1}(C)$ is closed, hence $f$ is continuous as preimage of closed sets under $f$ is closed.

Sarvesh Ravichandran Iyer
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We want to show that, if $V$ is an open subset of $Y$, then $f^{-1}(V)$ is open in $X$. In other words, we need to show that, for any $x \in f^{-1}(V)$, there is an open neighbourhood $U_x$ of $x$ with $f(U_x)\subseteq V$.

If you want to use the tube lemma, the key is to notice that if $y\neq f(x)$, then there is an open set $U_y\times V_y \subseteq X\times Y$ where $U_y$ is an open subset of $X$ containing $x$ and $V_y$ is an open subset of $Y$ containing $y$, which is disjoint from $\Gamma_f = \{(x,f(x)): x\in X\}\subseteq X\times Y$. Thus if $z\in U_y$, then $f(z)\notin V_y$. Now the open sets $\{V_y:y \in Y\backslash \{f(x)\}\}$ has $\bigcup_{y \in Y\backslash \{f(x)\}} V_y = Y\backslash \{f(x)\}$, so the $V_y$ are almost an open cover of $Y$.

The rest of the proof works by adding a suitable open set to the $\{V_y\}$ so that you obtain an open cover of $Y$, and then examining what compactness gets you.

You can also reorganise the above argument in the following way (based on the sort of approach Carsten S. takes abvoe) which perhaps makes the "tube" come to the forefront more. It at least separates clearly the roles played by compactness of $Y$ and the property that the graph is closed in proving the continuity of $f$.

Claim 1: If $Y$ is compact, then the map $p_1 \colon X\times Y \to X$ is closed (that is, if $F\subseteq X\times Y$ is closed, then its image $p_1(F)$ is closed.

Proof of claim: We show $p_1(F)^c\subseteq X$ is open in $X$. Indeed if $x \in p_1(F)^c$, then $(\{x\}\times Y) \cap F = \emptyset$, or equivalently $(\{x\}\times Y \subseteq F^c$, an open subset of $X\times Y$. Thus each $(x,y)\in \{x\}\times Y$ lies in an open of the form $U_y\times V_y \ni (x,y)$ with $U_y$ open in $X$ and $V_y$ open in $Y$. Then the $\{V_y: y \in Y\}$ form an open cover of $Y$, so that by compactness, there is a finite subset $B\subseteq Y$ with $Y = \bigcup_{b in B} V_b$. Since $x \in U_y$ for all $y \in Y$, $U_x := \bigcap_{b \in B} U_b$ is an open neighbourhood of $x$ in $X$ and $U_x\times Y\subseteq F^c$ (i.e. since $\{x\}\times Y$ is compact, if it is disjoint from $F$ then it lies in a tube $U_x\times Y$ which is also disjoint from $F$. But then clearly $U_x \subseteq p_1(F)^c$ and so $p_1(F)^c$ is open as required.

Claim 2: If $\Gamma_f= \{(x,f(x)): x \in X\}$ is closed in $X\times Y$ and $Y$ is compact, then $f$ is continuous.

(Sketch) proof: I won't give a complete proof of this so as not to give the game away on everything, notice there is a map $F\colon X \to X\times Y$ given by $F(x) = (x,f(x))$, and it is easy to see that $F$ is continuous if and only if $F$ is. Now the restriction $p_f$ of $p_1$ to $\Gamma_f = F(X)$ is the inverse of $F$, and $p_f$ is certainly continuous, so its inverse $F$ is continuous provided $p_f$ is an open map, that is, a map for which the image of an open set is open. Since $F$ and $p_f$ are bijections, it is easy to see that $p_f$ is an open map if and only if it is a closed map. But if $\Gamma_f$ is closed, then any closed set in $\Gamma_f$ is a closed set in $X\times Y$, and claim $1$ showed that such sets are sent to closed sets in $X$, hence $p_f$ is a closed (and open) map as required.

krm2233
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