$(X, \tau\_X) $ and $(Y, \tau\_Y) $ be two topological spaces.


$\forall f\in Y^X$ with $\text{Gr}(f) $ is closed implies $f\in C(X, Y) $.


Question : Does this implies $(Y, \tau\_Y) $ is compact?




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Notation:


$Y^X$: Set of all functions from $X$ to $Y$.


$C(X, Y) =\{f\in Y^X: f \text{ is continuous }\}$


$\text{Gr}(f) =\{(x,f(x)):x\in X\}\subset X×Y$




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Question $1$:(prove/disprove)


$(\forall X$ and $\forall f\in Y^X \text{ with } \textrm{Gr}(f) \subset X\times Y$ closed $\implies f\in C(X, Y) ) \implies Y$ is compact.


Question $2$:(prove/disprove)


For a fixed non discrete space $(X, \tau) $ and $\forall f\in Y^X$ having closed graph is continuous then $Y$ is compact.



(Dis)prove : if every function with closed graph are continuous then the target space is compact
Let me try to answer these questions under the assumption that $Y$ is $T\_1$.


We start with a set $E$ along with a filter $\mathcal F\in\mathcal P(E)$. We can then cook up a topological space $X$ with underlying set $\{x\}\sqcup E$ with topology given by the discrete topology $E^\delta$ on $E$, and $\{x\}\sqcup U$ for every $U\in\mathcal F$. It follows from definition that, for every topological space $Y$, a map $f\colon X\to Y$ is continuous if and only if $f\rvert\_E\colon E\to Y$ *tends to* $f(x)\in Y$ along the filter $\mathscr F$, namely, for every neighborhood $V$ of $f(x)\in Y$, there exists a set $U\in\mathcal F$ such that $f(U)\subseteq V$. In short, the continuity captures the convergence along the filter.


What about the graph of a map $f\colon X\to Y$ being closed? Here we need the assumption of $Y$ being $T\_1$, so that we only have to test the existence of open neighborhoods at every point $(x,y)\in X\times Y$ where $y$ runs through all points of $Y\setminus f(x)$. Unwinding the definitions, it is equivalent to saying that, there exists a subset $U\in\mathcal F$ and an open neighborhood $V$ of $y\in Y$ such that


1. $y\not\in V$; and
2. for every $x\in U$, we have $f(x)\not\in V$.


Equivalently, it is saying that $y$ is not a cluster point of $f$ along $\mathcal F$.




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A positive answer to Question 1. It is known that, a topological space is compact if and only if every filter on the underlying set admits a cluster point. We assume that $Y$ is non-compact, thus it is non-empty, and there exists a filter on $Y$ which does not admit any cluster point. We fix such a filter $\mathcal F$, an arbitrary point $y\in Y$, and let $E=Y$ be a set endowed with the filter $\mathcal F$. We now define $f\colon X\to Y$ by $f(x)=y$ and $f\rvert\_E=\operatorname{id}$. The previous analysis shows that $f$ is not continuous but its graph is closed.




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A sketch of a negative answer to the second question. More precisely, we have


**Proposition.** For *every* topological space $X$, there exists a non-compact but Hausdorff topological space $Y$ such that every map $X\to Y$ with closed graph is continuous.


Roughly speaking, the compactness of $Y$ can be more complicated than what a fixed topological space $X$ can see. To see this, we introduce a slightly "quantitative" version of compactness:


**Definition.** Let $\kappa$ be a strong limit cardinal (i.e., for every $\lambda<\kappa$, we have $2^\lambda<\kappa$). We say that a topological space $X$ is *$\kappa$-compact* if, for every set $E$ with $\lvert E\rvert<\kappa$ and every filter $\mathcal F$ in $E$ (note: $\kappa$ being strong limit implies that $\lvert\mathcal F\rvert\le 2^{\lvert E\rvert}<\kappa$), every map $E\to X$ admits a cluster point along $\mathcal F$.




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**Update:** This definition is equivalent, by the axiom of choice, to the following simpler one: every filter *base* $\mathcal F$ on $X$ with $\lvert\mathcal F\rvert<\kappa$ admits a cluster point in $X$. The usual proof shows that this $\kappa$-compactness is equivalent to the open-cover condition that every open cover of size $<\kappa$ contains a finite subcover.




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**Lemma.** Let $\kappa$ be a strong limit cardinal and $Y$ a $\kappa$-compact topological space. Then for every topological space $X$ with $\lvert X\rvert<\kappa$, the projection $X\times Y\to X$ is closed.


The usual proof leads to


**Corollary.** Let $\kappa$ be a strong limit cardinal and $Y$ a $\kappa$-compact topological space. Then for every topological space $X$ with $\lvert X\rvert<\kappa$, a map $X\to Y$ is continuous if and only if its graph is closed.


Now in order to see the Proposition above, it suffices to construct a $\kappa$-compact non-compact topological space for a chosen strong limit cardinal $\kappa>\lvert X\rvert$. If I am not mistaken, we can take an ordinal $\lambda$ with cofinality greater than $\kappa$, with the order topology.


