Can a closed interval be an open set in a topological space? In the exercises of the book 'Topology without tears', we are asked to prove that the following collection of subsets of $\mathbb{R}$ is a topology:
The set containing $\emptyset$, $\mathbb{R}$, and every closed interval $[-n,n]$, for $n$ any positive integer.
This does seem to work out to be a topology given that it satisfies all 3 axioms, but if so, then these sets $[-n,n]$, which are closed intervals become open sets on the topology, by definition. Is this correct? Isn't this contradictory?
 A: It is correct! However, while the intervals $[-n,n]$ are closed in the usual topology on $\Bbb R,$ which is why we often refer to them as "closed intervals," they are not closed in this one.
Another thing to keep in mind is the old saying among topologists: "Sets are not doors. They can be open, closed, both, or neither."
In this topology, $\emptyset$ and $\Bbb R$ are both open and closed. Each interval $[-n,n]$ is open but not closed, while each set $(-\infty,-n)\cup(n,\infty)$ is closed but not open. All other subsets of $\Bbb R$ are neither open nor closed under this topology.
Another example is the topology consisting of all subsets of $\Bbb R.$ This gives an instance where every subset of $\Bbb R$ is both open and closed.
A: Here are two reasons why it's not contradictory.
The shorter reason is that there are lots of topological spaces that have subsets that are both open and closed. (Indeed, the entire space and the empty set are always both open and closed. But there are lots of nontrivial examples as well, such as any subset of a discrete space.) Therefore, the fact that a subset is closed does not contradict its being open.
But the more essential reason is that a particular set can have lots of topologies placed upon it, and the sets that are open and closed in one of those topologies don't have to have anything to do with the sets that are open and closed in another topology. In this case, we are defining a new topology on $\Bbb R$, and so the open sets are, by definition, the ones listed, and the closed sets are precisely the complements of those open sets (so $\Bbb R$, $\emptyset$, and $(-\infty,-n)\cup(n,\infty)$ for every $n\in\Bbb N$). In particular, $[-n,n]$ is not a closed set in this new topology.
The language is definitely confusing, because the fact that we normally call $[-n,n]$ a "closed" interval is because it's closed in the usual topology on $\Bbb R$. (Well, the interval nomenclature probably arose earlier than the topological nomenclature, but never mind....) But that's just one way to refer to that particular set, not a guarantee of closedness in every situation. It might have been less confusing if the problem had said: "The set containing $\emptyset$, $\Bbb R$, and every interval $\{x\in\Bbb R\colon -n\le x\le n\}$ for $n$ any positive integer".
A: To complement the other answers, let me just say that the definition of topology is very broad, so it lends itself to all kind of weird counterexamples. For example, still on $\mathbb R$, consider this topology: 
$$
\mathcal T=\{\varnothing, \mathbb R, [0,5],\mathbb R\setminus[0,5]\}. 
$$
This is a topology, since it satisfies the 3 axioms. Now take any sequence $\{x_n\}$ with $x_n\in[0,5]$ for all $n$ (or, for $n$ sufficiently large). Let $t\in[0,5]$ be any number. Now, given any $V\in\mathcal T$ such that $t\in V$, we have that $V=\mathbb R$ or $V=[0,5]$. And then $x_n\in V$ for all $n$ (or, for all $n$ sufficiently large). It follows that $\lim_{n\to\infty} x_n=t$. So, for instance, the sequence $(2,2,2,\ldots)$ converges to $\pi$. So does the sequence $\{1/n\}_n$. Both also converge to $1.7$.
Even more disconcerting is that the above shows that any sequence that eventually stays either in $[0,5]$ or in its complement, converges to uncountably many different points. 
The intuition we bring from classical calculus works more or less fine with metric spaces, but not with the general idea of a topology. For a topology to start behaving in a more "natural" way, we usually need it be Hausdorff. 
