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Consider the set

$$ C = \{ x \in \mathbb R^n\colon Ax = b \}, $$ where $A$ is some matrix in $\mathbb R^{m \times n}$ and $b \in \mathbb R^m$ such that the system is feasible. I aim to prove such set is closed.

My attempt. When proving a set is closed, it suffices to show that $\overline C \subseteq C.$ Then, let us consider and arbitrary element $x \in \overline C.$ This means that there exists a sequence $(x_n)_{n \in \mathbb N} \subset C$ such that $x_n \to x$ as $n \to \infty.$ Since $x_n \in C,$ for every $n \in \mathbb N$, we can say that $$ A x_n = b.$$ Letting limits as $n \to \infty$ on both sides, it yields $$ \lim_{n \to \infty} Ax_n = \lim_{n \to \infty} b \Leftrightarrow A(\lim_{n \to \infty} x_n) =b \Leftrightarrow Ax = b, $$ proving that $x \in C,$ just by definition of $C$. Therefore, $C$ is closed.

Is this enough to show closedness of $C$? How would one reproduce the same proof using the usual epsilon-delta definition of sequence convergence? Thanks for any help in advance.

xyz
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1 Answers1

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It is enough. Why would you prefer an epsilon-delta proof? (it would be less readable).

You can even use a smarter argument: $C=A^{-1}(\{b\})$ and $\{b\}$ is closed and $A$ is continuous, hence $C$ is closed.

However, here is a detailed epsilon-delta proof:

Since we want to prove that $\Bbb R^n\setminus C$ is open, let us choose some $x$ in it and prove that for some $r>0$ the open ball $B(x,r)$ is disjoint from $C.$

Let $\varepsilon=\|Ax-b\|.$ If we find some $r>0$ such that $$\|v\|<r\implies\|Av\|<\varepsilon,$$ we will be able to conclude: $$y\in B(x,r)\implies\|y-x\|<r$$$$\implies\|b-Ay\|=\|(b-Ax)-(Ay-Ax)\|$$$$\ge\|b-Ax\|-\|Ay-Ax\|=\varepsilon-\|A(y-x)\|>0$$$$\implies y\notin C.$$

The final hit is to deduce such an $r$ from the existence of an $M>0$ such that $\forall v\in\Bbb R^n\quad\|Av\|\le M\|v\|$: take $r=\varepsilon/M.$

Anne Bauval
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