5
$\begingroup$

I wrote a proof, and I just wanted to verify if it is correct.

Proof : Suppose not. Then, $[0,1] = U\cup V$ for $U$ and $V$ open in $[0,1]$ such that $U\cap V = \emptyset$. Let $0\in U$. Consider $s= sup\{t \mid [0,t] \subset U\}$. Clearly, $s\in[0,1]$.

If $s\in U$, then since $U$ is open, $\exists \epsilon >0$ such that $B(s;\epsilon) \subset U$ . But then $s+\epsilon \in U$ , and $s+\epsilon > s$, contradicting the definition of $s$. So, $s \notin U$.

Then, $s\in V$. But again, since $V$ is open, $\exists \epsilon >0$ such that $B(s;\epsilon)\subset V$. But, since $s$ is the least upper bound of $U$, for this $\epsilon$, $\exists s_{1} \in U$ such that $s_{1} \in B(s;\epsilon)$. Then, $s_{1} \in U\cap V$, which contradicts the fact that $U$ and $V$ are disjoint.

So, our assumption is wrong and $[0,1]$ is connected.

$\endgroup$
  • $\begingroup$ Looks good to me. $\endgroup$ – Arthur Nov 6 '18 at 14:19
2
$\begingroup$

You're probably not looking for the answer still after a year, but for those who come across this in the future, it may be helpful. You are pretty much right about your proof, but here is a more formal way of writing it.

That is that "If $a<b$, then the subspace $[a,b]$ of $E^1$ is connected".

Let $a<b$ and let $[a,b]$ be a subspace of $R$ with the $E^{1}$ topology. For the sake of contradiction, assume that $[a,b]$ is not connected. Then $[a,b]= U \cup V$, where $U$ and $V$ are nonempty disjoint open sets in $[a,b]$. Without loss of generality we may assume that $b \in V$. Because $U$ is nonempty and bounded above (by $b$), the axiom of completeness states that $U$ has a least upper bound $s$. We will prove that $s$ is not an element of either $U$ or $V$, and this will yield a contradiction. Suppose $s \in U$. Since $U$ is open, there exists an $\epsilon> 0$ such that $B_{\epsilon}(s) \subseteq U$. Thus, since $s+ \frac{\epsilon}{2} \in U$ and $s<s+ \frac{\epsilon}{2}$, there exists an element in $U$ that is greater than $s$. Therefore, $s$ is not an upper bound for $U$. This is a contradiction. Thus, $s \not \in U$. Now suppose $s \in V$. Then, because $V$ is open, there exists an $\epsilon>0$ such that $B_{\epsilon}(s) \subseteq V$. Then $s- \frac{\epsilon}{2}$ is an upper bound for $U$. But $s- \frac{\epsilon}{2}<s$, which is a contradiction because $s$ is the \emph{least} upper bound for $U$. So $s \not \in V$. Thus, $s \not \in U \cup V$, which is a contradiction. Therefore, $[a,b]$ is connected.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think it would be clearer for the second part, if you added the argument that if $s-\frac{\epsilon}{2}$ is not an upper bound for $U$, then there must be a $p\in U$ such that $s-\frac{\epsilon}{2} < p$ and since $s=\sup{U}$, $p\in(s-\frac{\epsilon}{2},s)\subseteq B_{\epsilon}(s) \subseteq V$ which cannot be the case since $U$ and $V$ are disjoint. $\endgroup$ – üzeyir Jun 10 at 16:43
1
$\begingroup$

TheoremA topological space $X$ is connected iff every continuous function $f: X \to\{±1\}$ is a constant function.

ProofLet $X$ be a connected space and $f: X \to \{±1\}$ a continuous function. We want to show that $f$ is a constant function. If $f$ is non-constant, then it is onto. Let $A = f^-{1}(1)$ and $B = f^{-1}(-1)$. Then $A$ and $B$ are disjoint non-empty subsets of $X$ such that $A$ and $B$ are both open and closed subsets of $X$ and $X = A U B$.(Why?). This is a contradiction. Therefore $f$ is constant.

Conversely, let us assume that $X$ is not connected. Therefore there exist two disjoint proper non-empty subsets $A$ and $B$ in $X$ such that $A$ and $B$ are both open and closed in $X$ and $X = AU B$. Now we define a map $f: X \to \{±1\}$ as

$f (x) = \begin{cases} 1,&x\in A\\-1,& x \in B\end{cases} $$ $.

Then $f: X \to\{±1\}$ is a continuous non-constant function. (Why?) This completes the proof.

Now I am giving you a shorter proof of your problem.

Suppose $j$ is any interval if $f: J \to\{±1\}$ is an onto Continuous function, then there exist $x,y\in J$ such that $f(x) = -1$ and $f (y) = 1$. By the intermediate value theorem, there exists $z$ between $x$ and $y$ such that $f(z) = 0$, a contradiction to our assumption that $f$ takes only the values $\pm 1$. Hence no such $f$ exists and hence $J$ is connected. It follows that any interval is connected.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm sorry but I couldn't really follow the proof. What is $f$ here? And why showing that no onto continuous function from an interval J to $\{\pm 1\}$ exists shows that $J$ is connected? $\endgroup$ – P-addict May 13 at 11:53
  • $\begingroup$ I have edited. Now you can see. $\endgroup$ – Ryszard Ebgelking May 13 at 12:30
  • $\begingroup$ Here $j$ is any interval. $\endgroup$ – Ryszard Ebgelking May 13 at 12:32
  • $\begingroup$ Nice answer, thanks! $\endgroup$ – P-addict May 13 at 20:22

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.