Show that the relation $R$ is closed if and only if $R^{-1}$ is closed

Question

Let $S$ and $T$ be Banach spaces and consider a relation, $R\subseteq S\times T$ in their direct sum. We call a relation closed if $R$ is closed as a subset of the direct sum $S\times T$. We define the inverse relation by $$R^{-1}=\{(t,s)\in T\times S:(s,t)\in S\times T\}\subseteq T\times S.$$

I want to show that $R$ is closed if and only if $R^{-1}$ is closed. Here is my attempt:

First of all notice that the map $M:R\rightarrow R^{-1}$ which assigns $(s,t)\mapsto(t,s)$ is a bijection.

To see that it is a surjection, notice that for every $(t,s)\in R^{-1}$ we find $(s,t)\in R$ such that $M((s,t))=(t,s)$.

On the other hand, to show that it is an injection, consider $(s_1,t_2),(s_2,t_2)\in R$ and suppose that $M((s_1,t_1))=M((s_2,t_2))$. Then $$M((s_1,t_1))=M((s_2,t_2))\implies(t_1,s_1)=(t_2,s_2).$$

Suppose, then, that $R$ is closed in $S\times T$. Then, for every $(s,t)\in S\times T$ there is some sequence $((s_n,t_n))_n$ in $R$ such that $\lim_{n\to\infty}((s_n,t_n))_n=(s,t)$, which happens if and only if $\lim_{n\to\infty}s_n=s$ in $S$ and $\lim_{n\to\infty}t_n=t$ in T.

Then, $M((s_n,t_n))=(t_n,s_n)$ for every $n\in\mathbb N$. But then, we know that $\lim_{n\to\infty}t_n=t$ in $T$ and $\lim_{n\to\infty}s_n=s$ in $S$ so that $\lim_{n\to\infty}((t_n,s_n))_n=(t,s)$ in $R^{-1}$. Hence, $R^{-1}$ is closed in $T\times S$.

Here are my questions on the above:

1. Is what I have written for the injectivity claim correct? I have tried to proceed by considering $a,b\in X$ for which $f(a)=f(b)$ and showing that this implies $a=b$. Above, starting with $a:=(s_1,t_1)$ and $b:=(s_2,t_2)$ and considering their image under $f:= M$, I haven't exactly obtained that $a=b$, and so I am wondering whether what I have done is sufficient, or if I have overlooked anything.

2. I have used that $\lim_{n\to\infty}((s_n,t_n))_n=(s,t)$ if and only if $\lim_{n\to\infty}s_n=s$ in $S$ and $\lim_{n\to\infty}t_n=t$ in T. Is it possible to explicate the above to show that the map $M$ is continuous (in fact, an isometry) in order to pass to the limit in $M\lim_{n\to\infty}((s_n,t_n))_n=\lim_{n\to\infty}M((s_n,t_n))_n=\lim_{n\to\infty}((t_n,s_n))_n=(t,s)$ in $T\times S$? Is it the case that one uses this fact without explicating it or not?

My thinking for the claim in 2. here is that $\|M(s,t)\|=\|(t,s)\|=\|(s,t)\|$ for all $(s,t)\in S\times T$.

Answer 1

I think your proof of bijectivity of $M$ is correct. You could explicate injectivity like this: $$M((s_1,t_1)) = M((s_2,t_2)) \iff (t_1,s_1) = (t_2,s_2) \iff t_1 = t_2 \; \wedge \; s_1 = s_2 \iff (s_1,t_1) = (s_2,t_2).$$

However, I see problems with your proof of closedness of $R^{-1}$.

Suppose, then, that $R$ is closed in $S×T$. Then, for every $(s,t)∈S×T$ there is some sequence $((s_n,t_n))_n$ in $R$ such that $\lim_{n→∞}((s_n,t_n))_n=(s,t)$.

This is actually saying that $R$ is dense in $S \times T$, not closed in $S \times T$.

You could proceed by considering $M$ on the whole space $S \times T$, instead of only on $R$, and showing that $M$ is a homeomorphism (i.e. continuous with continuous inverse). We will show that $M$ is continuous, the case of $M^{-1}$ is analogical.

We need to choose a norm on the product (they are equivalent) - let it be the maximum norm (i.e. $||(s,t)|| = \max \{||s||,||t||\}$).

For that let $(s_n,t_n)_n$ be a sequence in $S \times T$ that converges to a point $(s,t) \in S \times T$. This means that $$||(s_n,t_n) - (s,t)|| = \max \{||s_n-s||,||t_n-t||\} \rightarrow 0,$$ hence, $$||s_n - s|| \rightarrow 0; ||t_n-t|| \rightarrow 0.$$ But now $$||M((s_n,t_n)) - M((s,t))|| = ||(t_n,s_n) - (t,s)|| = \max \{||t_n - t||,||s_n - s||\} \rightarrow 0$$ and $M((s_n,t_n)) \rightarrow M((s,t))$. We showed that $M$ is continuous.

Now, $R^{-1} = M(R)$ is closed as it is a homeomorphic image of a closed set.

I would also like to mention that this statement remains true if $S,T$ are just topological spaces and we endow $S \times T$ and $T \times S$ with the product topology.