2
$\begingroup$

Let $A\subset R^{n}$ . Then $A$ is disconnected iff there exists a continuous and surjective function $f:A\to${0,1}

How can I prove this? To prove $\rightarrow$, I know that if $A$ is disconnected, then there are two open, non empty and disjoint sets $U,V\subset R^n$ such that $A=U\cup V$ , $A\cap U \ne \emptyset $, $A\cap V \ne \emptyset$ and $A \cap U \cap V=\emptyset$ .

Then we define a function $f(x)=\begin{cases} 0, & x \in U \\ 1, & x \in V\\ \end{cases}$

We can see that this function is surjective because both sets are non empty and there is not an $x$ such that $f(x)=0$ and $f(x)=1$ (the sets are disjoint). Is this correct?

Then I don't know how to prove the continuity of $f$, I've tried using the fact that $f$ is continuous iff the inverse image of an open(closed) set is open(closed) in $A$, but I still struggle using the concepts of relative open and closed sets(my professor didn't explain them very well) so I don't know how to finish the proof.

Any help will be apprecciated, thanks.

$\endgroup$
  • $\begingroup$ You should mention that $U,V$ are open $\textbf{relative to} A$. $\endgroup$ – Tim Raczkowski Dec 2 '15 at 2:29
2
$\begingroup$

The topology on $\{0,1\}$ is the discrete topology, i.e. all subsets are open.

Restrict $f$ to $A$. All you need to check that the preimage of each open set is open. It is clear for $\{0,1\}$ are $\emptyset$, and you have shown that $f^{-1}(\{1\})=V\cap A$ (which is open in the induced topology on $A$) and $f^{-1}(\{0\})=U\cap A$ (which is open in the induced topology on $A$). So you are actually done!

$\endgroup$
  • $\begingroup$ My professor didn't even cover the discrete topology (just a sophomore Vector Calculus class), could you explain me why every subset of {0,1} is open? Thanks $\endgroup$ – mobzopi Dec 2 '15 at 2:30
  • $\begingroup$ @mobzopi: Sure. The discrete topology is defined to be the topology where every subset is open. But if you are uncomfortable with abstract definitions, you can just think about it as the induced topology from the reals onto the subset {0,1} (in this case). $\endgroup$ – Kyle Dec 2 '15 at 2:36
0
$\begingroup$

Forget $\mathbb{R}^n$ and stay in the induced topology of $A$.

A space $A$ (by its own right) is connected iff (by definition) there does not exist two non-trivial, disjoint, proper, open subsets $U,V$ of $A$ such that $A=U \cup V$ .

Supposing $A$ is not connected, then there are such $U,V$. Your function $f$ defined in the way you did is continuous, because the pre-image of every open subset of $\{0,1\}$ (I assume you are considering the discrete topology) is either $\emptyset, A, U$ or $V$, hence open.

$\endgroup$
0
$\begingroup$

$'\implies '$

You have chosen the right function $f$ .To justify its continuity note that $\{0,1\}$ is equipped with discrete topology. So what are the open sets available ?

They are $\{0\},\{1\},\{0,1\},\emptyset $ Then $f^{-1}(\{0\})=U;f^{-1}(\{1\})=V;f^{-1}(\{0,1\})=A;f^{-1}(\emptyset)=\emptyset$ which is open in each case.Thus $f$ is continuous.

$\endgroup$

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.