7

Let A be a subset of $\Bbb R^n$ and let $\mathbf w$ be a point in $\Bbb R^n$. The translate of A by $\mathbf w$ is denoted $\mathbf w$ + A and is defined by $$\mathbf w+A \equiv \{\mathbf w + \mathbf u\mid \mathbf u\text{ in A}\}.$$ $\mathbf a.$ Show that A is open if and only if $\mathbf w$ + A is open.
$\mathbf b.$ Show that A is closed if and only if $\mathbf w$ + A is closed.
For part $\mathbf a.$ I know that if A is open then $\mathbf u$ is an interior point of A. Is it correct to say: $$ \therefore \mathbf w + \mathbf u \text{ must be an interior point of } \mathbf w + A \implies \mathbf w + A \text{ is open.}$$ and I'm not sure how to get started on $\mathbf b.$

Dan C
  • 95

2 Answers2

4

For part (a), it suffices to show that $B_r(w+a) = w + B_r(a)$. But this is easy since $$x\in B_r(w+a) \iff ||x-(w+a)||<r$$ $$\iff||(x-w)-a||<r$$ $$\iff x-w\in B_r(a)$$ $$\iff x\in w+B_r(a)$$

For part (b), just note that any set $S$ is closed iff $S^c$ is open. So, $$A\text{ closed } \iff A^c \text{ open }$$ $$\iff w+A^c \text{ open }$$ $$\iff (w+A)^c \text{ open }$$ $$\iff w+A\text{ closed } $$

MPW
  • 43,638
  • Hi @MPW, that is just for one inclusion, what about showing that $w + B_{r}(a) \subseteq B_{r}(w+a)$? – Overachiever Feb 05 '20 at 19:27
  • @Overachiever : It is not for just one inclusion. It is for both. Note these are double-headed arrows, so the implication chain is reversible. – MPW Feb 05 '20 at 19:35
  • @MPW If A is open and B is closed, is A+B both closed and open because of part (a) and (b)? – Mikalo Jan 19 '21 at 10:03
  • @Mikalo : No. The only sets that are both closed and open are $\varnothing$ and $\mathbb R^n$. The argument fails because $A+B= \bigcup\limits_{a\in A}(a + B)$. The union is the problem. – MPW Jan 19 '21 at 10:27
  • @MPW Why is the union the problem? Since the union of closed sets is closed right? – Mikalo Jan 20 '21 at 06:49
  • @Mikalo : No, only the union of finitely many closed sets is guaranteed to be closed. After all, every set $S$ is a union of closed sets since $S=\bigcup\limits_{s\in S}{s}$ and each ${s}$ is closed. – MPW Jan 20 '21 at 12:22
3

The function $\phi(x) = x+w$ is a homeomorphism (continuous with a continuous inverse) hence it maps open sets to open sets and closed sets to closed sets (since closed sets are the complements of open sets).

Note that $w+A = \phi(A)$.

copper.hat
  • 172,524