continuous functions between Topologies

**shmounal** · October 18th 2011, 06:34 AM

Hi there, I've got quite a big assignment about topological spaces and think I'm doing OK apart from this bit...

Let X and Y be topological spaces. Show that $f : X \rightarrow Y$ is continuous if and only if
$\forall A \subseteq X, f(\overline{A}) \subseteq \overline{f(A)}$ . By considering the map $f : \mathbb{R} \rightarrow \mathbb{R} , f(x) = x/(1 + x^2)$ , show that we

do not expect $f(\overline{A}) = \overline{f(A)}$ .

For the first part I know, f is continuous if and only if the image of the closure of every subset $\forall A \subseteq X$ is contained in the closure of the image which implies the result but don't think this is really me showing this, its just me stating a rule...

For second part I'd have thought we would expect $f(\overline{A}) = \overline{f(A)}$ so I'm obviously not understanding something properly.

Any help much appreciated!!

**Plato** · October 18th 2011, 07:12 AM

Originally Posted by shmounal

Let X and Y be topological spaces. Show that $f : X \rightarrow Y$ is continuous if and only if
$\forall A \subseteq X, f(\overline{A}) \subseteq \overline{f(A)}$ . By considering the map $f : \mathbb{R} \rightarrow \mathbb{R} , f(x) = x/(1 + x^2)$ , show that we
do not expect $f(\overline{A}) = \overline{f(A)}$ .
For the first part I know, f is continuous if and only if the image of the closure of every subset $\forall A \subseteq X$ is contained in the closure of the image which implies the result but don't think this is really me showing this, its just me stating a rule...
For second part I'd have thought we would expect $f(\overline{A}) = \overline{f(A)}$ so I'm obviously not understanding something properly.

I am confused by the first part. You say that you already know what you are asked to show. What is the sense in that?

For the second part I think the function should be $f(x)=\frac{x^2}{1+x^2}$ .
Then let $A=(0,\infty)$ . Compare $f(\overline{A})~\&~\overline{f(A)}$

**shmounal** · October 18th 2011, 11:49 AM

Well for the first part I'm basically just quoting out of the lecture notes we have been going over. I 'know' it because it's given as a rule but guess they're looking for me to actually prove it which I'm not sure on how to do! For the second part I can see it would make sense if that were the question but in both hand out and online the question we have is one stated. Worse comes to worse I'll just solve it for $f(x) = x^2/(1 + x^2)$ and just see what happens!

**Plato** · October 18th 2011, 12:16 PM

Originally Posted by shmounal

Well for the first part I'm basically just quoting out of the lecture notes we have been going over. I 'know' it because it's given as a rule but guess they're looking for me to actually prove it which I'm not sure on how to do! For the second part I can see it would make sense if that were the question but in both hand out and online the question we have is one stated. Worse comes to worse I'll just solve it for $f(x) = x^2/(1 + x^2)$ and just see what happens!

Many good topology textbooks, such as Helen Cullen's, give a discussion of alternate definitions of continuity . It appears that you are being asked to show that the statement is an alternate definition of continuity. What is you basic definition?

As for the function $f(x) = x/(1 + x^2)$ , it does not have the suggested property but $g(x) = x^2/(1 + x^2)$ does.
Recalling that $\mathbb{R}=\overline{\mathbb{R}}$
$f(\mathbb{R})=[-0.5,0.5]$ a closed set.

But $g(\mathbb{R})=[0,1)$ , not closed.

**shmounal** · October 19th 2011, 07:18 AM

Yeah can see that second part now, thanks!!

The definition for continuous we have is...

Let $(X, T)$ and $(Y, U)$ be topological spaces. A function
$f : X \rightarrow Y$ is continuous iff
$\mu \in U$ implies $f^{−1} (\mu) \in T$ .

That is, inverse images of open sets are open.

It even says lower down that the part they're asking about is just an alternate definition. I'm guessing you have to do some playing around with complements and things as thats what most of the stuff so far has been like but don't really know. Sorry, bit unclear but I don't really know what it wants myself!

**Plato** · October 19th 2011, 07:59 AM

Originally Posted by shmounal

The definition for continuous we have is...
That is, inverse images of open sets are open.

This Fred Croom's approach to this question.
Suppose that we have a function $f:X\to Y$ having the property that
$1)\;\left( {\forall A \subseteq X} \right)\left[ {f\left( {\overline {A\;} } \right) \subseteq \overline {f(A)\;} } \right]$

Show that 1) implies 2) where
$2)\;\left( {\forall C \subseteq Y} \right)\left[ {C{\text{ is a closed set in }Y} \Rightarrow {\kern 1pt} f^{ - 1} (C){\text{ is a closed set in }X}} \right]$

Now show that 2) implies your given definition.

The key to showing $1)\Rightarrow\;2)$ is this fact:
If $\overline{A}\subseteq A\text{ then }A\text{ is closed.}$

Hint start with $f\left( {\overline {f^{ - 1} (C)\;} } \right)$ .

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