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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.

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$X$ and $\varnothing$ are open sets in every topology that are both open and closed. – Enjoys Math Oct 15 '13 at 22:22
What you want to prove is that if $A$ is open then $X - A$ is closed and vise versa, in any topological space X. – Enjoys Math Oct 15 '13 at 22:23
The interesting thing is to prove that the real line is connected. – dfeuer Oct 15 '13 at 22:24
In the discrete topology every set is both open and closed, so you can't prove what you want without some restriction on your topology. – Mark Bennet Oct 15 '13 at 22:29
The question omits to say which space these sets are in. If it's the real line or any Euclidean space, then there are only two sets that are simultaneously open and closed: the empty set and the whole space. In spaces with more than one connected component, any union of connected components is about open and closed. – Michael Hardy Oct 15 '13 at 22:43
up vote 1 down vote accepted

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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There exists sets which are both open and closed.

For example: $\mathbb R$ with the standard topology.

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This could be a subtly misleading or confusing thing to say. Generally speaking, "open" and "closed" are not absolute properties of a "set" $X$; they are considered meaningful only relative to a given topology on a set $Y$ containing $X$ as a subset. If $X$ is considered a subset of itself, then of course $X$ is open and closed relative to any topology you choose to put on $X$, "standard" or otherwise. – user43208 Oct 16 '13 at 2:03
The concerned set is in $$ R^n $$ – Parikshit Khanna Oct 16 '13 at 4:30

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