[Math] Prove open set is not closed

analysisgeneral-topology

The question might sound ridiculous, but I am not able to prove it with rigor. I tried proving it by the following definitions ONLY.

Open set: A set $U$ is open if for every $a$ belonging to $U$, there is
some $r = r(a) > 0$ such that the ball $B_r(a)$ is contained in $U$.

Closed set: A set $A$ is closed if it contains all of its limit
points.

I tried using that $A$ closed and open would render $A'$ open and closed (provable using the above definitions) but that didn't lead me anywhere.

Best Answer

I believe that the essential question here is this: Why isn't any non-empty proper subset of $\Bbb R^n$ both open and closed? This is another way of saying that $\Bbb R^n$ is connected: there do not exist nonempty, disjoint open sets $A,B\subseteq \Bbb R^n$ such that $A\cup B=\Bbb R^n$ (that is, there do not exist $A$ and $B$ that separate $\Bbb R$.

It turns out that the toughest part is showing that $[0,1]$ is connected. This is done by using two facts:

  1. Every non-empty set of real numbers which is bounded above has a supremum.
  2. Between any two real numbers there is another real number.

The proof starts like this:

Suppose for the sake of contradiction that $A$ and $B$ separate $[0,1]$, and suppose WLOG that $0\in A$. Let $m=\inf B$. See if you can continue from there.

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