How to prove the preimage of a sequentially-closed set under a linear mapping is also sequentially-closed?

Question

Preimage is defined as $X = \{x \in \mathbb{R}^n:Ax \in S\}$, where $A$ is a linear mapping $A \in \mathbb{R}^{m,n}$, and $S \subseteq \mathbb{R}^m$ is a sequentially-closed set.

The definition of sequential-closedness is that $S$ is sequentially-closed iff every convergent sequence in $S$ has its limit in $S$.

What I tried: Suppose $X$ has a convergent sequence whose limit is not in $X$, then since the mapped sequence is in S, whose limit is not in $S$ (1). This is contradicted to what we assume that $S$ is a closed set, so all convergent sequences in $X$ has their limits in $X$, which proves that $X$ is a closed set.

Now my question is how can I prove (1) is true?

Answer 1

Suppose $\{x_k\}_{k\in\mathbb{N}}$ is a sequence in $X$, converging to $x\in X^c$. We will show that $Ax_k$ converges to $Ax$. Let $\{e_1,..., e_n\}$ be the standard basis for $\mathbb{R}^n$ and $\|\cdot\|$ denote the Euclidian norm, then:

$$\|Ax_k-Ax\|=\|A(x_k-x)\|=\|\sum_{i=1}^n(x_{ik}-x_i)Ae_i\| \leq \sum_{i=1}^n|x_{ik}-x_i|\|Ae_i\| \leq \max_i\|Ae_i\|\sum_{i=1}^n|x_{ik}-x_i|$$ $$\leq \max_i\|Ae_i\| \: \sqrt{n} \: \|x_k-x\|$$ where the last line follows from Cauchy-Schwarz inequality. Hence $\|Ax_k-Ax\| \rightarrow0$. Since $Ax \in S^c$, $\{Ax_k\}_{k\in\mathbb{N}}$ is a sequence in $S$, whose limit is not in $S$

Answer 2

You attempted a proof by contradiction. There's nothing wrong with taking this approach.

You correctly started from a sequence $x = (x_1, x_2, \dots)$ in $X$, which you assumed to be convergent to a limit, call it $L$, that does not lie in $X$. So far, so good. Let's stop for a brief moment to make a note of the assumption that we are making, that $$ L\text{ does not lie in }X. \tag{*}\label{star} $$

You correctly realized that the contradiction we aspire to arrive at, is that the sequence $Ax = (Ax_1, Ax_2, \dots)$ is a convergent sequence in $S$, whose limit does not lie in $S$.

You couldn't figure out how to show that $Ax$'s limit does not lie in $S$. However, you seem to have missed the point that you additionally need to show that $Ax$ is convergent.

The reason why $Ax$ is convergent derives from the fact that the mapping $f:\mathbb{R}^n \rightarrow \mathbb{R}^m$ that assigns to every $v \in \mathbb{R}^n$ the value $f(v) := Av$, is continuous (it is continuous since it is linear, as all linear mappings are continuous). In fact, $Ax$'s limit is $AL$: $$ AL = f(L) = f\left(\lim_{n\rightarrow\infty} x_n\right) \overset{\text{continuity}}{=} \lim_{n\rightarrow\infty} f(x_n) = \lim_{n\rightarrow\infty} Ax_n. $$

So we wish to show that $AL$ does not lie in $S$, and, indeed, it can't lie in $S$, since, if it did, than $L$ would have been in $S$'s preimage, $X$, in contradiction to assumption \eqref{star}.