Why is empty set an open set? I thought about it for a long time, but I can't come up some good ideas. I think that empty set has no elements,how to use the definition of an open set to prove the proposition.
The definition of an open set: a set S in n-dimensional space is called open if all its points are interior points.
 A: It's the union over the empty subfamily of the topology: that's all.
A: A set is open or closed (or neither) inside another set (actually a set, equipped with a topology).
By the definition of topology (https://en.wikipedia.org/wiki/Topology#Mathematical_definition) the empty set is always open and its complement (i.e. the whole other set) is always closed.
Therefore is the emtpy set open and closed in every topology.
A: Here's another perspective:
Suppose the empty set is not considered open.
Consider the sine function $\sin\colon \Bbb R \to \Bbb R$.
Then $(3\,.\,.\,4)$ is open (presumably), but $\sin^{-1}(3\,.\,.\,4)=\varnothing$ is not open, so the sine function is not continuous.
We don't want that!
Another thing: one of the key properties of open sets is that the intersection of any two open sets is open. If the empty set were not considered open, then that wouldn't be true anymore.
A: My answer would have nothing to compare with @HagenvonEitzen's wit and to the point, but here's a way to think of it:
The complement of an empty set is the whole set, which of course contains everything including all limit points. Hence the whole set is closed, and therefore it compliment, empty set is open.
Hagen's argument can be made to show empty set is closed and whole set is open. Because all points in empty sets are limit points, so empty set is closed. So its compliment, whole set is open.

Another (better) way to think of this:
A topological space generalizes the concept of a metric space. With this view, a function is continuous iff the inverse image of an open set is open. Since a function that maps the entire space onto a single point is always continuous, the empty set is open. Take an open set which does not contain the single point. Its inverse image is the empty set.
Above is a proof for the definition, however, empty set is open by the definition of a topology. But it's good to know why things are defined the way they are.

A final note: It should be "the" empty set, since it is unique.
A: It's vacuously true. All points $\in\emptyset$ are interior points, have a blue-eyed pet unicorn, and live in Surrey.
A: A different way to think of why this absolutely has to be true in spaces with some disjoint open sets (which is definitely the case in $\mathbb R^n$) is that the intersection of any finite collection of then must be open. Clearly in these spaces the empty set arises in this way. This is motivation, not a complete argument. The fact that $\emptyset=\bigcup_{A\in\emptyset}A$ is a better reason.
Note that these are general topological answers, not really tied to real spaces.
A: Because  it is the compliment of a closed set.
A: The way I wrapped my head around it, is by defining a subset $E$ of a metric space $M$ to be open if $\forall x: x\in E \rightarrow x\text{ is an interior point of }E$. If $E$ denotes the empty set $\emptyset$, no $x$ is contained in $E$, so the implication is automatically true for all $x$.
A: Indeed, all the points of the empty set are interior points.
This is a statement that is ''vacuously true".
In other words: If you produce an element of the empty set, I will be happy to show that it is an interior point. This can be done because first of all you can't produce one. Compare to Russel's assertion in an apocryphal tale, "Given an inconsistent proposition, I can prove any statement"(paraphrased).
A: Another way to think about it is that, since the empty set has no points, you can't find a point in the empty set that is NOT an interior point to violate the definition. So it is true "by default," or vacuously.
A: The empty set is open, by definition. Nothing more to it. 
A: Consider the following definition of an open set: For every element in A, there exists a ball of the element contained in A. What is the negation of that statement? Well, it is: There exists an element in A such that every ball is not contained in A. Clearly, the empty set does not satisfy the negation of an open set, therefore it must be vacuously true that the empty set satisfies the definition of an open set.
