I'm studying General Topology by the book of James Munkres and there he defined connectedness this way:

"Let X be a topological space. A separation of X is a pair U, V of disjoint nonempty open subsets of X whose union is X. The space X is said to be connected if there does not exist a separation of X."

and then he says that "if X is connected, so is any space homeomorphic to X" and this implication it's an equivalent definition, but it doesn't clear for me why. I thought prove this implication by contrapositive, but I don't have idea how to prove this. Someone can help me to understand why?

Thanks in advance!

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    $\begingroup$ Can you quote what he actually says? $\endgroup$ – B. Pasternak Apr 17 '16 at 16:35
  • $\begingroup$ I edited my topic $\endgroup$ – George Apr 17 '16 at 16:35
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    $\begingroup$ I read the thing you are referring to in his book, and he doesn't say anything about "an implication being an equivalent definition". He says that connectedness is a topological property, which it of course is, and a first argument he gives is that it is formulated entirely in terms of open sets. Then he mentions that it is a property preserved by homeomorphisms, which is a different way of saying that something is a topological property. $\endgroup$ – B. Pasternak Apr 17 '16 at 16:40
  • $\begingroup$ I understood now, I was thinking that is an equivalent definition of connectedness, thanks! $\endgroup$ – George Apr 17 '16 at 16:46

Let's prove "if $X$ is connected, then so is any space homeomorphic to $X$".

So let $X$ be connected, and let $Y$ be homeomorphic to $X$, which means there is a function $h: X \rightarrow Y$ that is 1-1, onto (so a bijection) such that $h$ and $h^{-1}$ are both continuous.

Suppose (for a contradiction) that $Y$ is not connected, so there exists a separation $U,V$ of $Y$. So $U,V$ are both non-empty and open in $Y$, disjoint and their union is $Y$.

Then by continuity of $h$, $h^{-1}[U]$ and $h^{-1}[V]$ are open, and as (for all functions) $\emptyset = h^{-1}[U \cap V] = h^{-1}[U] \cap h^{-1}[V]$, these sets are also disjoint, and also $X = h^{-1}[U \cup V] = h^{-1}[U] \cup h^{-1}[V]$, their union is $X$. Both are non-empty as $h$ is surjective.

So $h^{-1}[U],h^{-1}[V]$ are a partition of $X$ which cannot be, as $X$ is connected. Contradiction.

Note that we only need $h$ onto and continuous. Being 1-1 and the inverse being continuous are not needed. In fact, we showed that the continuous image of a connected space is connected, which is uusally one of the first things to prove about connectedness.

The remark from Munkres only meant to say, I think, that connectedness is entirely defined in terms of open sets and set theory (disjointness, union) and so will be a so-called topological property, so any space homeomorphic to it also has it. E.g. metric completeness is not such a property. I don;t think it was meant as an alternative definition at all, just a quick remark about the property being clearly a topological one.


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