Let us consider $$\mathcal X = \{(a,b)|a \in A, \ b = f(a)\}, $$ where $A \subset L^1(\mathbb R)$ has empty interior and $f:L^1 \to L^1$ is a bijective map. Does $\mathcal X$ also have empty interior? My intuition would suggest yes, but I don't have a proof. Bonus question: If yes, can the assumption on $f$ be relaxed? If no, can the assumption on $f$ be strengthened to make the result true?
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6$\begingroup$ $X$ will have empty interior as a subset of $L^1\times L^1$, or $A\times L^1$ if you wish, for any subset $A$ and any function $f$. This is clear by contrapositive - if the interior was nonempty, then it would contain two different points $(a,b),(a,b')$. $\endgroup$– WojowuCommented Aug 10, 2022 at 18:32
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$\begingroup$ @Wojowu What do you mean? $\endgroup$– RikuCommented Aug 10, 2022 at 20:44
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If the interior of $X$ were nonempty there would be nonempty open sets $U$ and $V$ in $L^1$ such that $U\times V\subseteq X$. But then $U\subseteq A$ shows that $A$ would have nonempty interior.
Note however that the graph of a function hardly ever has nonempty interior: in the situation above we would have $(a,f(a))\in U\times V$ for some $a\in A$. Because $f$ is a function this implies that $V=\{f(a)\}$, and so $f(a)$ must be an isolated point of $L^1$, and $f$ would be constant on the open set $U$. So, if your range space does not have isolated points a set like $X$ always has empty interior.
