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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?

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  • $\begingroup$ 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
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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).

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  • $\begingroup$ 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

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