complement of compact set is connected Let A be a compact subset of R, the real numbers. Prove that the complement of A in the complex numbers C is connected. 
My thoughts: If A is compact then it is contained in a finite union. So if it's complement in C was disconnected it would imply C was disconnected-contradicton
 A: My thoughts:
Whenever possible, I prefer dealing with path-connected spaces to connected spaces, because I can more easily visualize it.  If $A \subset \mathbb{R}$ is compact, then there's an $R > 0$ such that $A \subset [-R, R]$.  Now if we have two points $z, w \in \mathbb{C} \setminus A$, then we have a couple of easy cases:


*

*If only one of the two points happens to lie on $\mathbb{R}$ (say it's z), we can move vertically from $z$ to $\Re z + \Im W$ and then move along that line to $w$.

*If they happen to have the same nonzero imaginary part, then just connect them linearly.

*If they happen to have differing nonzero imaginary parts, just move horizontally (left or right) until you're passed $[-R,R]$, move vertically to the right half plane, and then you're free to move linearly.

*If they both happen to lie on $\mathbb{R}$, move vertically from one point into a half plane, move horizontally until you have the appropriate real coordinate, and then move vertically again.
A: Be careful. It isn't clear what you mean by "a finite union." Also, your reasoning doesn't work--the complement of $\Bbb R$ in $\Bbb C$ is disconnected, even though $\Bbb C$ isn't.
Here's an alternative approach: Suppose $A\subseteq\Bbb R$ such that $\Bbb C\setminus A$ is disconnected. Note that $\Bbb C\setminus\Bbb R$ is contained in $\Bbb C\setminus A,$ so the upper open half-plane must lie in a connected component of $\Bbb C\setminus A,$ as must the lower open half-plane. Use this to show that $A=\Bbb R.$ Hint: If not, then there is some real $z$ lying in $\Bbb C\setminus A.$ Now, consider the following (which is a good exercise to prove):

Lemma: If $X$ is a connected subset of a topological space, and $\overline X$ denotes its closure, then for any set $Y$ with $X\subseteq Y\subseteq\overline X,$ we have that $Y$ is connected.

In particular, since the upper open half-plane $H^+$ is connected, then $H^+\cup\{z\}$ is connected. Likewise, $H^-\cup\{z\}$ is connected, so since the union of two nondisjoint connected sets is connected, then $(\Bbb C\setminus\Bbb R)\cup\{z\}$ is connected. Since the closure of this set is all of $\Bbb C,$ then the result follows from the Lemma. (Do you see how?)
Hence, we have a stronger result: If $A$ is any proper subset of $\Bbb R$ (in which compact subsets of $\Bbb R$ are included), then its complement in $\Bbb C$ is connected.

P.S.: It is tempting to try to proceed as in your (now-deleted) answer below, and say that if $\Bbb C\setminus A$ were disconnected, then it is the disjoint union of two non-empty closed sets $E$ and $F,$ and so $\Bbb C$ is the disjoint union of the three non-empty closed sets $E,F,$ and $A,$ and so we reach the contradictory conclusion that $\Bbb C$ is disconnected.
The problem with that approach is that we must distinguish between sets that are (relatively) closed in $\Bbb C\setminus A$ and sets that are closed (in $\Bbb C$). The particular case $A=\Bbb R$ shows us how this reasoning falls apart. Indeed, $\Bbb C\setminus A$ is the disjoint union of $H^+$ and $H^-,$ both of which are non-empty and (relatively) closed. Since $\Bbb R$ is closed (in $\Bbb C$), it would seem by the reasoning above--without the parenthetical remarks--that $\Bbb C$ is the disjoint union of three closed non-empty sets. But neither $H^+$ nor $H^-$ is closed in $\Bbb C,$ so those distinctions are critical!
