Let $f:(M_1,d_1)\to (M_2,d_2)$ be a mapping between two metric spaces.

a)Let $A\subseteq M_1$ be open and $B\subseteq M_1$ closed. Show through the use of counterexamples that in general $f(A)$ is not open and $f(B)$ is not closed in $M_2$.

b) Show that f is continuous when the pre-images of closed sets in f are closed again.

Just started learning about metric spaces in general and this one piqued my interest in the textbook.

Here's what I thought:

a) Although I'm allowed to use counterexamples I don't know how to start. I was thinking that I need to find a set in $M_1$ whose ball is not open in $M_2$, right? And similarly for B, that the corresponding set in $M_2$ is not closed, meaning that the complement is not open?

But how can I show that? I mean, I have no explicit norm given to set up the balls. Any ideas?

b) We weren't at the chapter of continuity in metric spaces so I tried reading about it somewhere. It seems kind of similar to proving continuity in $\mathbb{R}$? But again, I don't know how to approach this without a norm. According to my textbook I'm supposed to show that

$d_2(f(x),f(x_0))<\epsilon$ if $d_1(x,x_0)<\delta$, correct?

But I seriously don't know how to use that to deal with b), especially since they require pre-images of f.

Any tips or hints would be greatly appreciated.

Edit: My approach for a): $$ d(x,y) := \begin{cases}0, &x = y\\1, & x\neq y\end{cases} $$

$B_r(x)=\{y\in \mathbb{R}|d(x,y)<r\}$ $x\in \mathbb{R},~r>0$

$B_{1/2}(0)=\{0\}$ is open for $x=y$.

For $d_2:|x-y|$

No finite number of open intervals so that the intersection is zero.

$\rightarrow f(B_{1/2}(0))$ is not open.

For the closed case:

You said I should take the complement of all the balls used for the open case, meaning:

$B_r(x)=\{y\in \mathbb{R}|d_1(x,y)>r\}$ for $x\in\mathbb{R},r>0$.

$\rightarrow$ $B_{1/2}^{d_1}(0)={1}$ is closed in $(\mathbb{R},d_1)$?

So for $d_2(x,y):=|x-y|$ and $f:(\mathbb{R},d_1)\to (\mathbb{R},d_2),x\mapsto x$. So similarly it should be $f(B_{1/2}^{d_1}(0))=\{1\}$, and $\{1\}$ is not closed because for all $r>0:B_r^{d_2}(0)={0}$ and not $\{1\}$, right?

  • $\begingroup$ writing $B_r(x)=\{y\in\mathbb R|d_1(x,y)>r\}$ is misleading. IMHO usually one expects $B_r(x)$ to be the ball around $x$, not its complement. $\endgroup$ – luckyrumo Jun 21 '15 at 14:02

a) In such exercises, you are allowed to use any metric and mapping you wish to. A nice one for counterexamples is $$ d(x,y) := \begin{cases}0, &x = y\\1, & x\neq y\end{cases} $$ What is the ball $B_r(x)=\{y\in\mathbb R\mid d(x,y)<r\}$ around $x\in\mathbb R, r>0$ with this metric? Can you find a mapping that maps it to a non-open set? (Remember that $d_1$ and $d_2$ may be different.)

(Side Remark: A metric space is by no means normed, for example if it has the metric $d$ given above.)

b) Usually continuous is defined as "the preimages of open sets are open". To prove your statement, use that open sets are the complements of closed ones:

Let $f:S\to T$ be a function where preimages of closed sets are closed. Let $A\in T$ be open, $B:=T\setminus A$. Hence B is closed, and by assumption $\tilde B=f^{-1}(B)$ is closed to. Use that to show that $\tilde A=f^{-1}(A)$ is open, which concludes your proof.

My suggestions for your approach above

$$ d_1(x,y) := \begin{cases}0, &x = y\\1, & x\neq y\end{cases} $$

Hence in $(\mathbb R, d_1)$ the open balls are $B_r(x)=\{y\in \mathbb{R}|d_1(x,y)<r\}$ for $x\in \mathbb{R},~r>0$

$\Rightarrow A:=B^{d_1}_{1/2}(0)=\{0\}$ is open in $(\mathbb R, d_1)$.

Let $d_2(x,y):=|x-y|$ and $f:(\mathbb R,d_1)\to(\mathbb R,d_2), x\mapsto x$

Thus $f(A)=\{0\}$, but $\{0\}$ is not open in $(\mathbb R, d_2)$ since $\not\exists r>0:B_r^{d_2}(0) = \{0\}$

For the closed set case, it suffices to take the complements of all the balls above:

Let $B := A^C = \mathbb R\setminus \{0\}$. $B$ is closed in $(\mathbb R, d_1)$ as complement of the open set $A$.

$\Rightarrow f(B) = B$ is not closed in $(\mathbb R, d_2)$ since its complement $A$ is not open (see above).

  • $\begingroup$ About a): So, if I were to pick $r=0.5$ the ball would be open for $x=y$, but otherwise the points would jump out of the ball? About b): I don't know how to show continuity yet. I just don't know how to start. $\endgroup$ – Fabian Henry Jun 21 '15 at 9:11
  • $\begingroup$ @Fabian: a) Exactly, $B_{1/2}(0) = \{0\}$ with the $d$ above. Now think about a metric space where that set is not open, and use the identity map. b) I'll edit my answer $\endgroup$ – luckyrumo Jun 21 '15 at 9:15
  • $\begingroup$ How can I use an identity map on metrics? Is it similar to functions? $\endgroup$ – Fabian Henry Jun 21 '15 at 9:18
  • $\begingroup$ @Fabian: If $M_1=M_2=\mathbb R$, $d_1=d$ (from above) and $d_2$ any metric on $\mathbb R$, you can use the identity $\mathbb R\to\mathbb R$. The metrics have nothing to do with it. $\endgroup$ – luckyrumo Jun 21 '15 at 9:21
  • $\begingroup$ Alright, I tried the same approach as you showed in b) with the mappings. And for $d_2$ I chose $|x-y|$. So, I try to get $f(B_{1/2}(0))$, right? So how can I show that $B_{1/2}'$ is also not open like $B_{1/2}(0)$? $\endgroup$ – Fabian Henry Jun 21 '15 at 9:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.