show that continuous maps preserve connectedness

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July 25th 2010, 01:03 PM
hatsoff

show that continuous maps preserve connectedness

Hi guys.

Suppose we have two metric spaces $X$ and $Y$ , and we have a function $f:X\to Y$ which is continuous on $X$ . If $S$ is a connected subset of $X$ , then it's easy enough to show that $f(S)=\{f(x):x\in S\}$ is connected.

However, suppose there is a function $g:X\to Y$ which is only continuous on $S$ , but not necessarily the rest of $X$ . How do we show that $g(S)$ is connected?

Any help would be much appreciated. Thanks!
July 25th 2010, 01:31 PM
tonio

Quote:

Originally Posted by hatsoff

Hi guys.

Suppose we have two metric spaces $X$ and $Y$ , and we have a function $f:X\to Y$ which is continuous on $X$ . If $S$ is a connected subset of $X$ , then it's easy enough to show that $f(S)=\{f(x):x\in S\}$ is connected.

However, suppose there is a function $g:X\to Y$ which is only continuous on $S$ , but not necessarily the rest of $X$ . How do we show that $g(S)$ is connected?

Any help would be much appreciated. Thanks!

Sketch: suppose $g(S)=U\cup V\,,\,U,V\in Y$ open

$\Longrightarrow S\subset g^{-1}(U)\cup g^{-1}(V)\,,\,g^{-1}(U)\,,\,g^{-1}(V)$ open in $S$ (why)? By connectedness of S one of

these two last sets must be empty, so...

Tonio
July 25th 2010, 01:36 PM
hatsoff

Thanks, but in your sketch you assume that $g(S)$ is open, which may not be the case. If on the other hand we say simply that $g(S)\subseteq U\cup V$ , then I won't be able to show that $g^{-1}(U)$ and $g^{-1}(V)$ are open.

Or am I misunderstanding something---always a possibility!---?
July 25th 2010, 01:42 PM
tonio

Quote:

Originally Posted by hatsoff

Thanks, but in your sketch you assume that $g(S)$ is open, which may not be the case.

I didn't mean but that's what must be deduced from my typos. I assume $g(S)$ that it is the union (oops! and I should have written disjoint union) of two open sets in $g(S)$ (another typo here), which means both $U,V$ are the intersection of $g(S)$ with open subsets of $Y$ , and NOT necessarily open in $Y$ . Sorry for that.

Tonio

If on the other hand we say simply that $g(S)\subseteq U\cup V$ , then I won't be able to show that $g^{-1}(U)$ and $g^{-1}(V)$ are open.

Or am I misunderstanding something---always a possibility!---?

.
July 26th 2010, 01:56 PM
hatsoff

So does anyone else have an idea?

I was thinking that maybe we could show that for any function $g:X\to Y$ , if the restriction of $g$ to $S\subseteq X$ is continuous, then there is a function $h:X\to Y$ which is continuous on $X$ , and whose restriction to $S$ is equal to that of $g$ . In that case, I could show that $S$ is connected implies $h(S)=g(S)$ is connected. However, I do not know how to prove that the continuous restriction of a function to $S$ may be expanded to a function continuous on $X$ . Heck, I'm not even sure it's true!

Help would be much appreciated!
July 26th 2010, 02:19 PM
Dinkydoe

Assuming S is nonempty,

By contraposition we can assume $G(S)= U\cup V$ where $U,V\subset Y$ open, nonempty and disjoint. (That is, assuming G(S) is not connected)

Then as Tonio said, $f^{-1}(U), f^{-1}(V)$ are open disjoint and nonempty, and $S= f^{-1}(U)\cup f^{-1}(V)$ ......and what does this mean?
July 26th 2010, 04:09 PM
hatsoff

If $g(S)$ is disconnected, then we cannot assume that $g(S)=U\cup V$ for disjoint open sets $U$ and $V$ . Rather, we can only assume that $g(S)\subseteq U\cup V$ for such sets. And since $g$ is only known to be continuous on $S$ , then that means $g^{-1}(U)$ and $g^{-1}(V)$ may be nonopen.

Thanks for the attempt though!
July 26th 2010, 04:32 PM
Defunkt

Quote:

Originally Posted by hatsoff

If $g(S)$ is disconnected, then we cannot assume that $g(S)=U\cup V$ for disjoint open sets $U$ and $V$ . Rather, we can only assume that $g(S)\subseteq U\cup V$ for such sets. And since $g$ is only known to be continuous on $S$ , then that means $g^{-1}(U)$ and $g^{-1}(V)$ may be nonopen.

Thanks for the attempt though!

How did you define connectedness? The definition I'm familiar with says that a space $X$ is disconnected if there exist two disjoint open sets $U,V$ such that $X=U \cup V$ ..

Also, you can simply define $h = g|_S$ and then $h:S \to Y$ is continuous and thus $h(S) = g(S)$ is continuous
July 26th 2010, 04:38 PM
hatsoff

Quote:

Originally Posted by Defunkt

How did you define connectedness? The definition I'm familiar with says that a space $X$ is disconnected if there exist two disjoint open sets $U,V$ such that $X=U \cup V$ ..

$g(S)$ is not a space. It's just a subset of the space $Y$ .

My definition is as follows: A set $S$ is disconnected iff there are disjoint open sets $U,V$ covering $S$ with $U\cap S\neq\emptyset\neq V\cap S$ . It is connected iff it is not disconnected.
July 26th 2010, 04:39 PM
Defunkt

Quote:

Originally Posted by hatsoff

$g(S)$ is not a space. It's just a subset of the space $Y$ .

My definition is as follows: A set $S$ is disconnected iff there are disjoint open sets $U,V$ covering $S$ with $U\cap S\neq\emptyset\neq V\cap S$ . It is connected iff it is not disconnected.

What do you mean? Then how can we give it the subspace topology?

EDIT: I mean, how do you define a space? And then why is $g(S)$ not one?
July 26th 2010, 06:02 PM
Dinkydoe

Quote:

If is disconnected, then we cannot assume that for disjoint open sets and

I don't know what definition of connectedness you're using.... But a topological space is connected if it can not be written as a disjoint union of 2 non-empty open sets U,V. Thus not connected means we can find such U,V.
July 26th 2010, 06:23 PM
tonio

Quote:

Originally Posted by hatsoff

If $g(S)$ is disconnected, then we cannot assume that $g(S)=U\cup V$ for disjoint open sets $U$ and $V$ . Rather, we can only assume that $g(S)\subseteq U\cup V$ for such sets. And since $g$ is only known to be continuous on $S$ , then that means $g^{-1}(U)$ and $g^{-1}(V)$ may be nonopen.

Thanks for the attempt though!

It seems like you haven't yet fully understood the definition of connectedness, but there is another equivalent definition which can be helpful here:

A set $S$ in a top. space $X$ is disconnected iff there exists a continuous function from $X$ onto $\{0,1\}$ , where this last space is the

usual two-points one with the inherited topology from the usual topology in $\mathbb{R}$ .

Let us go back now to your problem: if $g(S)$ is disconnected there exists $h:g(S)\rightarrow \{0,1\}$ continous and onto, but then
$h\circ g:S\rightarrow \{0,1\}$ is cont. and onto, contradiction.

Tonio
July 27th 2010, 02:08 PM
hatsoff

I understand the definition of connectedness which I related earlier. Although I haven't gone and proved it, I can see how your alternate definition might be equivalent.

That said, I figured out how to prove the contrapositive: If $g(A)$ is disconnected, where $g:X\to Y$ is continuous on $A$ , then there are disjoint open sets $U,V$ covering $g(A)$ with $g(A)\cap U$ and $g(A)\cap V$ nonempty. Putting $U'=\bigcup\{B(d(x,A\cap f^{-1}(U))/2,x):x\in A\cap f^{-1}(U)\}$ and $V'=\bigcup\{B(d(x,A\cap f^{-1}(V))/2,x):x\in A\cap f^{-1}(V)\}$ satisfies disconnectedness for $A$ .

Thanks for the help, guys!