# Proof that $(0,1)$ is connected.

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 $s=sup(A)$.

• For clarification, did you mean connected or not connected? You said one in the title and another in the question statement. – Zéychin May 7 '13 at 14:41
• But, what if $s=1$ ? – mercio May 7 '13 at 14:51
• connected ofcourse! Sorry, don't know how that "not" ended up in that sentence. – harajm May 7 '13 at 14:52
• 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 $\mathbb R$. – Santiago Canez May 7 '13 at 14:53
• 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 $\mathbb R$ – Thomas Andrews May 7 '13 at 15:22

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.