Why is infinite intersections of all intervals $(-1/n, 1/n)$, where $n$ is positive integer, not open? I'm a non-mathematician who just started studying topology, hence if the question sounds naive pardon me.  
In Wikipedia article on "Open Set" it is said that:  

Note that infinite intersections of open sets need not be open. For example, the intersection of all intervals of the form $(−1/n, 1/n)$, where $n$ is a positive integer, is the set $\{0\}$ which is not open in the real line.

Would you please explain why $\{0\}$ is not open? Which of the axioms or properties of open set does it violate?
 A: the definition of an open set in a metric space $M$ is that for all elements x of the set, you can find a distance $\epsilon > 0$ such that $B(x; \epsilon)$ is contained in the set, where $B(x; \epsilon)$ =  $\{y \in M:\ d(x, y) < \epsilon\}$ is an $\epsilon$-open ball centered at $x$.
Here we are working with $R$, and $0$ is the only element of $\{0\}$, so the question is: Can we find an $\epsilon > 0$ such that $B(0; \epsilon)$ is contained in $\{0\}$? or in other words, the only element of $B(0; \epsilon)$ is $0$?. The answer is no. Take any $\epsilon > 0$, the open ball would be the interval $(0 -\epsilon, 0+\epsilon) = (-\epsilon, \epsilon)$. Since the real numbers are dense, you can always find an element in that interval, for example $0 \neq \epsilon/2 < \epsilon$. So we can't find such $\epsilon$ and it follows that the set is not open. By the same reasoning, any singleton in $R$ is not open
A: Well if you are studying topolgy you might know that we need to define a topolgy $\tau$ over a set ${X}$ here $X=\mathbb{R}$. So in certain case the singleton {0} can be an open set. For example in the discret topology, recall that the discret topology is $\tau_d=\{A \mid A \subset \mathbb{R} \}$. Hence since {0} is a subset of $\mathbb{R}$ then {0} $\in \tau_d $ so it is an open set in this topology. 
But of course in you example I suppose that $\mathbb{R}$ is endowed with the euclidean topology. 
To be short the topology is the set 
$\tau_{euclidean}=\{\bigcup_{i\in I} U_i\ \mid U_i=(a_i,b_i), a_i \lt b_i, \forall i \in I\}$ where $I$ is any set empty or coutable or uncountable.
So we say a set $A$ is open if and only $A \in \tau_{euclidean}$
Now for your question since {0} can not be written as union of open intervall then {0} is not open.
You can also prove that {0} is not open by showing that it the set is closed recall that a set is closed if the complementary is open. Since the only closed and open set in this topology is the all set ($\mathbb{R}$) and $\emptyset$ then {0} can not be open.
How to show {0} is closed 

$\mathbb{R} \setminus$ {0} $= \bigcup_{i=1}^{\infty} (-i,0) \cup(0,i)$ so then set is open so by definition {0} is closed 

A: The only sequence of elements in $\{0\}$ is defined by $$(\forall n\in \Bbb N) \;\; x_n=0$$
this kind of sequence converges to $0$ which is in $\{0\}$. thus this set is closed.
If it was open, its complementary  $\; (-\infty,0)\cup (0,+\infty) \; $ will be closed, which is not the case since the sequence $\frac 1n$ of its elements converges outside this set.
A: Note that in a Hausdorff space, finite-point sets are closed. So in particular $$\bigcap_{n=1}^\infty \left(-\frac1n,\frac1n\right)= \{0\}$$ is closed. Now, $\mathbb R$ is connected, so the only subsets of $\mathbb R$ that are both open and closed are $\mathbb R$ and $\varnothing$. It follows then that $\{0\}$ is not open.
