Why are the empty set and the set of all real numbers both open and closed?
sorry! am not clear with these questions
why an empty set is open as well as closed?
why the set of all real numbers is open as well as closed?
By definition, a set $A$ of real numbers is open when the following condition is met: $$ \hbox{$\forall x\in A, \exists\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subset A$,} $$ where $(a,b)$ denotes the open interval $\{x\in{\Bbb R}\,|\,a<x<b\}$. It should be not hard to convince yourself that the subsets $A=\emptyset$ and $A={\Bbb R}$ satisfy this condition.
Then, remember that $$ \hbox{$A$ is open $\iff {\Bbb R}\setminus A$ is closed} $$ again by definition. You conclude since $\emptyset={\Bbb R}\setminus{\Bbb R}$ and ${\Bbb R}={\Bbb R}\setminus\emptyset$.
Well the definition of a topological space $X$ specifies that both $X$ and the empty set must be open sets (if the topology is defined in terms of closed sets rather than open sets, it will stipulate that they are closed). But then it is just by definition that it must be open (or closed).
Then a set $A$ is said to be closed if and only if its complement $X - A$ is open. So if you look at the empty set its complement is $ X - \emptyset = X$ and $X$ is open by definition. Therefore the empty set is closed.
This should be pretty obvious. Take $\mathbb{R}$ (together with its equipped topology) for example. We have:
1. Since finite intersection of two open sets is open, it follows that $(1, 2) \cap (3, 4) = \emptyset$ must be open; The complement of $\emptyset$, which is $\mathbb{R}$, must be closed;
2. Since any union of two open sets is open, it follows that $ (- \infty, 1) \cup (-1, + \infty) = \mathbb{R}$ is open; By the same complement rule again, the complement of $\mathbb{R}$, which is $\emptyset$, must be closed.
Hope this helps.
In my opinion, the other answers to this question are quite poor, as they just cite the definition of a topology which indeed states that the whole space and the empty set are open and closed. The natural question that then follows is: why define it like that?
The idea of topology comes from metric spaces (of which $\mathbb{R}$ is an example), and one approach is the following: In metric spaces we can talk about convergence of sequences (If you know what convergence means in $\mathbb{R}$ you can follow this explaination). In mathematics we often talk about subsets being closed under a certain operation, for example, a subvectorspace has to be closed under addition. That means that if you take to vectors in that subspace, their sum has to be in that subspace aswel. Closed in metric spaces means the same for convergence. A subset $A$ in a metric space is called closed if it is closed under the operation of taking limits. i.e.: $$ (\forall n: a_n \in A) \implies \lim_n a_n \in A $$ (If the limit exists). It is obvious that both the empty set and the whole space satisfy this (can you see this?) so they are both closed. Now open can be defined as being the complement of a closed set and you're question is answered.
The idea of general topology is to then give an abstract generalization of these closed sets (or of the open sets) that works better, for example, it allows you to take products of spaces which in the metric spaces causes problems. This abstract generalization is made by observing some properties of open and closed sets in metric spaces, and then calling them a definition. One of these properties is that the empty set and the whole space are open and closed. So to conclude, the general definition of a topology is great, but not to give an (insightful) answer to this question