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We all know the classical proof that a path-connected topological space $X$ is also connected. I will recall it briefly, so that we all talk about the same thing.

Let $X$ be a path-connected topological space and assume for a contradiction that $X=A \cup B$, where $A \cap B = \emptyset$ and $A,B \neq \emptyset$ (i.e. $X$ isn't connected). Choose $a \in A$, $b \in B$ Let $\gamma: [0,1] \rightarrow X$ be a path in $X$ such that $\gamma(0)=a$, $\gamma(1)=b$. Then by continuity of $\gamma$ there is a decomposition $$[0,1]=\gamma^{-1}(A) \cup \gamma^{-1}(B)$$ and this decomposition implies that $[0,1]$ is not connected, which is not true. Hence we conclude that $X$ cannot be path-connected, if it isn't connected.

Is there a proof, which does not use the connectedness of $[0,1]$? I would like a slick short proof similar to the one above, if you know one. Or is there a good reason, why the connectedness of $[0,1]$ should be essential?

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Why do you know that $[0,1]=\gamma^{-1}(X)?$ – user May 28 '15 at 4:44
up vote 11 down vote accepted

The connectedness of $[0,1]$ is clearly crucial, as may be seen by considering what would happen if we used something else.

Let $K$ be any space with two distinguished points $p$ and $q$. Say that a space $X$ is $K$-connected if for any $x,y\in X$ there is a continuous map $f:K\to X$ such that $f(p)=x$ and $f(q)=y$. If $K$ is not connected, there is clearly no reason to expect $K$-connectedness of $X$ to imply connectedness of $X$. As an extreme case, let $K=\{0,1\}$ with the discrete topology: then every space is $K$-connected.

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Thank you very much for this enlightening answer! – Nils Matthes Apr 4 '12 at 8:54

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