# Path connected in the set of complex numbers

may i ask for a little help about a proof i have to show.

Let $U\subset \mathbb{C}$ be an open and path connected set. Show that if $W\subset U$ closed and open not empty subset, then $U = W$.

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(In other words, you want to show that $U$ is connected.) – Zev Chonoles Feb 23 '13 at 21:55
If i show that every path connected set is connected (what is actually a theorem) will be enough? – Lullaby Feb 23 '13 at 21:57
Yeah, should be. – Ben Millwood Feb 23 '13 at 22:03
The openess of $U$ has nothing to do here. – Damian Sobota Feb 23 '13 at 22:12

Pick $x \in W$ and $y \in U$. The goal is to show that $y \in W$. But there is a path from $x$ to $y$, say $p : [0,1] \to U$ with $p(0) = x$, $p(1) = y$. But then the image of $p$ is a connected set (because continuous image of a connected set) containing $x$ and $y$, so they must belong to the same connected component of $U$. But the connected component of $x$ is contained in $W$. So $x$ and $y$ are both in $W$.

But $y$ was an arbitrary member of $U$!

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Do you know that if $X\subseteq [0,1]$ is closed and open, then $X=[0,1]$ or $X=\emptyset$?

If so, assume $\emptyset \subsetneq W\subsetneq U$. That means there is some $x\in W$ and some $y\in U\setminus W$.

Consider a path $\gamma$ from $x$ to $y$, and specifically $\gamma^{-1}(W)$.

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I assume you know that $[0,1]$ is connected and a continuous image of a connected space is still connected.

Assume $\emptyset\neq W\neq U$. Then $U\setminus W$ is also closed and open in $U$. Take $x\in W$ and $y\in U\setminus W$. There is a path $\varphi:[0,1]\to U$ such that $\varphi(0)=x$ and $\varphi(1)=y$. But $W\cap\varphi([0,1])$ and $(U\setminus W)\cap\varphi([0,1])$ are closed and open and nonempty in $\varphi([0,1])$. What does it mean for $\varphi([0,1])$?

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Suppose $\rm W$ is closed and open, then $$\rm U = W \cup ((\mathbb C \backslash W) \cap U)$$

is a union of two disjoint non empty open sets, which is not possible because $\rm U$ is connected.

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I think the point of the problem is to show that path-connected implies connected. – Thomas Andrews Feb 23 '13 at 22:01

Since U is path-connected set, then it is connected, i.e. The only clopen subsets of U are itself and (fi) But W is a clopen non empty subset of U Then W = U...

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