2
$\begingroup$

I wasn't completely satisfied with the proof given by my professor that the open interval $(0,1)$ is connected in the subspace topology of Euclidian space. So I came up with this argument instead, please tell me if there is anything unclear about this proof. Am I using any unexplicit assumptions?

A proof by contradiction, assume that $(0,1)$ is disconnected. Let $A,B$ be disjoint non-empty open sets such that $X=A\bigcup B$. Then $A,B$ are also closed in $(0,1)$. Since $A=\{x\in (0,1) :x\in A\}$ we have that A is bounded by $1$, since $A$ is non-empty by assumption, the $LUB$ axiom tells us that $sup(A)=s$ exists. Furthermore $s\in \bar{A}$, since for all $r> 0$ there exists $x\in A$ such that $s-r<x<s$ because othervise $s-r$ would be the supremum of $A$. Hence $B_r(s)\bigcap A\neq \emptyset$, this shows that $s\in \bar{A}$. Since $A$ is closed $A=\bar{A}$ and hence $s\in A$. Since $A$ is open $B_{\epsilon }(s)\subseteq A$, i.e. $(s- \epsilon , s + \epsilon ) \subset A$, this implies that $s + \frac{\epsilon }{2} \in A$, contradicting the assumption that $s=sup(A)$.

$\endgroup$
  • 1
    $\begingroup$ For clarification, did you mean connected or not connected? You said one in the title and another in the question statement. $\endgroup$ – Zéychin May 7 '13 at 14:41
  • 1
    $\begingroup$ But, what if $s=1$ ? $\endgroup$ – mercio May 7 '13 at 14:51
  • $\begingroup$ connected ofcourse! Sorry, don't know how that "not" ended up in that sentence. $\endgroup$ – harajm May 7 '13 at 14:52
  • 1
    $\begingroup$ Your mistake is in the final sentence when you say "Since $A$ is closed, $A = \overline A$ and hence $s \in A$." We only know that $A$ is closed in $X$, but what you're claiming would require that $A$ be closed in $\mathbb R$ since the supremum $s$ only a priori exists in $\mathbb R$. $\endgroup$ – Santiago Canez May 7 '13 at 14:53
  • $\begingroup$ In the two topologies of $(0,1)$ and $\mathbb R$, $\bar A$ means different things. It is possible for $\sup A$ to not be in $A$. For example, if $A=[1/2,1)$, $\bar A=A$ in $(0,1)$ but $\bar A=[1/2,1]$ in $\mathbb R$ $\endgroup$ – Thomas Andrews May 7 '13 at 15:22
2
$\begingroup$

Your proof that $s\in\operatorname{cl}A$ is slightly flawed. At that point in the argument you don’t know that for each $r>0$ there is an $x\in A\cap(s-r,s)$; you know only that there is an $x\in A\cap(s-r,s]$. That is, $s$ could in principle be the supremum of $A$ not because it’s a limit from the left of elements of $A$, but simply because it’s the maximum element of $A$. An example would be the set $\left[0,\frac12\right)\cup\left\{\frac34\right\}$. Of course we know that an open set $A$ can’t really look like this, but you’d have to make a separate argument to show that $s$ is not just $\sup A$, but actually $\sup(A\setminus\{s\})$.

However, this flaw doesn’t affect the conclusion that $s\in\operatorname{cl}A$: if $s\in A$, then automatically $s\in\operatorname{cl}A$, and if $s\notin A$, then your argument works.

The other problem is that $s$ might be $1$. In that case you could replace $A$ with $B$ and go for the same basic argument, but what if $\sup A=\sup B=1$? To finish your proof, you’d have to show that that’s impossible, but I think that you’ll find that a bit tricky.

$\endgroup$

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.