Absolute values of a closed set's elements

Question

Let $A \subset \mathbb{R}^n$ be a closed set. Is the set $\{|x| : x \in A \} \subset \mathbb{R} $ closed?

Based on my intuition, I strongly believe that the above set is closed, but I couldn't give a precise proof. Can anyone help me?

Thanks!

Answer 1

It is closed.

Given $y=\lim y_n, y_n\in\{|x|:x\in A\}$ we have to show that $y\in\{|x|:x\in A\}$.

For each $y_n$ there is at least one $x_n\in A$ such that $|x_n|=y_n$. So consider a sequence $x_n$ with this property.

$\{x_n\}$ is bounded (why?) so there exists a subsequence $x_{n_k}$ such that $x_{n_k}\rightarrow x$ for some $x\in\mathbb{R}^n$ (Bolzano–Weierstrass theorem).

We know $x\in A$ because $A$ is closed. Moreover, $|x|=\lim |x_{n_k}|=\lim y_{n_k}=y$, so $y\in\{|x|:x\in A\}$.

Answer 2

Let $B=\{\|x\|:x\in A\}.$ Suppose $b\in \bar B.$

There is a sequence $\beta =(b_m)_{m\in \Bbb N}$ of members of $B$ with $b=\lim_{m\to \infty}b_m.$

And there is a sequence $\alpha=(a_m)_{m\in \Bbb N }$ of members of $A$ with $\|a_m\|=b_m$ for each $m\in \Bbb N.$

Now $\beta$ is a convergent series in $\Bbb R$ so the set $\{b_m:m\in \Bbb N\}$ is a bounded subset of $\Bbb R.$ So there exists $M\in \Bbb R^+$ such that $|a_m|=b_m\le M$ for each $m\in \Bbb N.$

So $\{a_m:m\in \Bbb N\}\subseteq A\cap \{x\in \Bbb R^n: \|x\|\le M\}=^{def}C.$

Now $C$ is a closed bounded subset of $\Bbb R^n$ (because $A$ is closed) so $C$ is compact. And $\alpha$ is a sequence of members of $C.$ So $\alpha$ has a subsequence $(a_{m_j})_{j\in \Bbb N}$ that converges to some $a\in C.$ And $a\in A$ (because $a\in C\subseteq A)$.

We have $|\;\|a\|-b_{m_j}\;|=|\;\|a\|-\|a_{m_j}\|\;|\le \|a-a_{m_j}\|$ and we have $\lim_{j\to \infty}\|a-a_{m_j}\|=0.$ Therefore $$b=\lim_{j\to \infty}b_{m_j}=\|a\|\in B.$$ So $b\in \bar B\implies b\in B.$ That is, $B$ is closed.

Remark: This will not work if you replace $\Bbb R^n$ with an arbitrary complete normed linear space, where a closed bounded subset may fail to be compact. E.g. if $A=\{(1+2^{-n})e_n:n\in \Bbb N\}$ where $\{e_n:n\in \Bbb N\}$ is an orthonormal Hilbert-space-basis for (infinite-dimensional) Hilbert space $H,$ then $A$ is closed in $H$, but $\{\|a\|:a\in A\}=\{1+2^{-n}:n\in \Bbb N\}$ is not closed in $\Bbb R.$