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Let $U$, $V$ be two open disjoint sets over $X$ such that $X=U\cup V$. Let $p\in U$. For any $q\in V$, let $\alpha :[0,1]\rightarrow X$ be a path connecting $p$ and $q$. Then the set $A:=\alpha^{-1}(U)$ is open in $[0,1]$. Let $x:=\sup A$. Then $y=\alpha (x)\in \partial U$:

  • If $W$ is an open set with $\alpha(x)\in W$, then $\alpha^{-1}(W)\cap A\neq \emptyset \implies \alpha(\alpha^{-1}(W)\cap A)=W\cap \alpha(A)=W\cap U\neq \emptyset$.
  • Likewise, $\alpha^{-1}(W)\cap ([0,1]-A)\neq \emptyset\implies \alpha(\alpha^{-1}(W)\cap ([0,1]-A))=W\cap (\alpha([0,1])-U)\subset W\cap V\neq \emptyset$.

Now, if $\alpha(x)\in V$, then $U\cap V=\emptyset \implies \alpha(x)\notin \partial U$ (which is a contradiction). Then $\alpha(x)\in U$. For every $y\in \partial U$, we can find a path (through path concatenation) such that $\alpha (x)=y$ for some $x\in [0,1]$, so $y\in U \implies U=\overline U$. That is, $U$ is open and closed in $X$. Then $\alpha^{-1}(U)$ must be open and closed in $[0,1]\implies \alpha^{-1}(U)=[0,1]$ (because $[0,1]$ is connected). Thus $q\in U$ for every $q\in V$, which is a contradiction, and $V$ must be empty. Therefore, $X$ is connected.

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  • $\begingroup$ what does $\alpha^{-1}\cap A$ mean? $\endgroup$ – yohBS May 29 '16 at 21:07
  • $\begingroup$ sorry, wrong typesetting. Already corrected. $\endgroup$ – Enric May 29 '16 at 21:08
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    $\begingroup$ What exact definition of "connected" are you using? Once you have that $\partial U$ is not empty (because it contains $y$), you're pretty much done. $\endgroup$ – David Schneider-Joseph May 29 '16 at 21:08
  • $\begingroup$ I'm using this: $X$ is connected iff $X=U\cap V$ (with $U$, $V$ disjoint and open) $\implies$ $U=\emptyset$ or $V=\emptyset$. $\endgroup$ – Enric May 29 '16 at 21:10
  • $\begingroup$ I think your argument is a very complicated way of saying that if $U$ and $V$ are open an disjoint then $\alpha^{-1}(U)$ and $\alpha^{-1}(V)$ are open and disjoint, and since they are a cover of $[0,1]$ it contradicts the connectedness of $[0,1]$. $\endgroup$ – yohBS May 29 '16 at 21:12
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You know that $[0,1]$ is connected, so the image of $\alpha$ is connected. On the other hand, $$ \alpha([0,1])=\bigl(\alpha([0,1])\cap U\bigr)\cup\bigl(\alpha([0,1])\cap V\bigr) $$ is written as the disjoint union of two relative open sets. Connectedness forces either $\alpha([0,1])\cap U=\emptyset$ or $\alpha([0,1])\cap V=\emptyset$. This is a contradiction, because $\alpha(0)=p\in U$ and $\alpha(1)=q\in V$.

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