Closure of the graph of a function Let $X_1, X_2$ be non-empty sets and let $R\subseteq X_1\times X_2$ such that for all $x\in X_1$ there is $y\in X_2$ such that $(x,y)\in R$.
Are there topologies $\tau_i$ on $X_i$ for $i=1,2$ and a continuous function $f:X_1\to X_2$ such that $R$ is the topological closure of $\text{graph}(f)$, where $\text{graph}(f) = \bigl\{\bigl(x,f(x)\bigr) \, : \, x\in X_1\bigr\}$?
 A: Take $X_1=X_2=\{1,2,3\}$ and $R=\{(x,y):x\neq y\}$. Then the diagonal is open as a complement of the closed subset $R$. As an open subset, the diagonal is a union of products $A_i\times B_i$ of some open subsets $A_i\subseteq X_1$ and $B_i\subseteq X_2$. But these must be singletons so both $X_i$'s are discrete. Therefore any graph of a function is closed - a contradiction since $R$ is not a graph.
A: According to https://mathoverflow.net/a/8994/4600,
$\log_2$ of the number of topologies on $n$ elements is $\sim n^2/4$.
So let's say there are $2^{n^2/4}$ topologies.
Then the number of topologies on $X_1=[n]=\{0,\dots,n-1\}$ and $X_2=[n]$ is
$$
2^{n^2/4}\cdot 2^{n^2/4} = 2^{n^2/2}.
$$
Consider now sets $R$ which contain $(k,0)$ for each $k<n$. There are
$$
2^{n(n-1)}
$$
such sets. But there are only
$$n^n = 2^{n\log_2 n}$$
functions, so the total number of $(\tau_1,\tau_2,f)$ is only
$$
2^{n^2/2}2^{n\log_2 n} < 2^{n(n-1)}
$$
for large $n$.
So there are just too many sets $R$.

Edit:
This is rather complicated compared to Adam Przeździecki's solution, but it does have the virtue of showing that "closure of" could be replaced by any other method of uniquely obtaining $R$ from $f$, $\tau_1$, and $\tau_2$.
