Closed graph theorem: Continuous with Hausdorf codomain ⟹ closed graph

Question

I have an answer to this question which does not use the Hausdorff property, but I suspect there is an issue with my solution opposed to the question. I am aware that there are already solutions to this question (e.g. here ) but I am not quite able to understand the posted solution in full. I also understand that the result can be shown by utilising the the fact that the diagonal is closed and that projections onto components are continous

The question stated formally is:

Let $X, Y$ be topological spaces and have $f: X \mapsto Y$ a set map. Let us define $G(f) = \{(x,f(x)) \in X \times Y | x \in X\} $ as the graph of $X$. Then show that if f is continuous and Y hausdorff then $G(f)$ is closed.

Definition of closure I use is the following:

Closure point of a subset $A$ is any point $x$ with the property that any nbhd $V$ of $x$ has $V \cap A \neq \emptyset$.

My solution:

Let $x^* = (x,f(x))$ be an arbitrary point in $G(f)$. If we consider $V$ in the nbhd of $f(x)$, then continuity gives us that $f(U) \subset V$ where $U$ is in the nbhd of $x$.

Then denoting $U^* = U \times V$, we have $U^* \cap G(f) = (U \times V) \cap G(f) \subset (U \times f(U)) \cap G(f) \neq \emptyset $

Since $U^*$ is a nbhd of $x^*$ and the choice of $x^*$ was arbitrary, we can conclude that $G(f)$ is closed.

Answer 1

You already seem to have gotten satisfactory answers to your concerns with your proof attemts, so instead here's an alternative solution if you're familiar with some of the basics of nets.

Let $(x,y)\in\overline{\operatorname{graph}(f)}$. Then there exists a net $\{(x_\alpha,f(x_\alpha))\}_{\alpha\in I}$ in $\operatorname{graph}(f)$ with limit $(x,y)$. As projections are continuous, we then have that $x_\alpha\to x$ and $f(x_\alpha)\to y$. Furthermore, as $f$ is continuous, $f(x_\alpha)\to f(x)$. As $Y$ is Hausdorff, nets in $Y$ have unique limits, and so $y=f(x)$. Consequently $(x,y)=(x,f(x))\in\operatorname{graph}(f)$, proving the claim.

Answer 2

Let $x \times y \in {\rm Cl} G (f)$ be arbitrary. We show that in fact $x \times y \in G (f)$, and since $x \times y$ was arbitrary, $G (f)$ must be closed.

Note that $x \times y \in G (f) \Longleftrightarrow y = f (x)$ by definition of "graph". Suppose for the sake of contradiction that $y \ne f (x) = z$. Since $Y$ is Hausdorff, there exist disjoint neighborhoods $U$ of $y$ and $V$ of $z$. Since $f$ is continuous, $f^{-1} (V)$ is a neighborhood of $x$. Since $f^{-1} (V) \times U \subseteq X \times Y$ is a neighborhood of $x \times y$ (by the definition of the product topology), it intersects $G (f)$ by hypothesis, say at a point $x' \times f (x')$. The first factor shows $x' \in f^{-1} (V) \implies f (x') \in V$, and the second factor shows $f (x') \in U$, which imply $f (x') \in U \cap V$. This contradicts the fact that $U$ and $V$ are disjoint.

Hausdorffness is required here. Consider for example $X = \{ 0, 1 \}$ in the discrete topology and $Y = \{ 0, 1 \}$ in the indiscrete topology. Define $f: X \to Y$ by $f (x) = 0$, which is continuous. The graph $G (f) = \{ 0 \times 0, 1 \times 0 \}$ is not closed in $X \times Y$. You can check this: every neighborhood of the point $0 \times 1 \notin G (f)$ contains the point $0 \times 0 \in G (f)$.