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?
 A: 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
A: 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. 
A: 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
${\Bbb R}={\Bbb R}\setminus\emptyset$.
A: 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.
A: $\emptyset$ is open. By definition, $X$ is open if $\forall x\in X$, there is a open set $U\subset X$ such that $x\in U$. So there is not any point in $\emptyset$, the condition of the definition is automatically satisfied (a logical convention).
$\mathbb{R}$ is open (check Andrea's answer), so its complement, $\emptyset$ is closed.
Therefore, $\emptyset$ is both open and closed.
A: By definition, a set $A$ of real numbers is open when the following condition is met:
(Note that this applies equally well to the set of real numbers, just substitute $A = R$.)
$$
\hbox{$\forall x\in A, \exists\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subset A$.}
$$
By the definition above, the set of all $x$ is no other than the set $A$ itself, and $A$ is already defined above to be open.
As for why an empty set is also open, is a rather tricky business AND DEFINITELY NOT IMMEDIATELY CLEAR why it is so as many people seem to say. First of all, it is defined rather than concluded from some logical reasoning. The reason we want an empty set to be made open is:

*

*We want to maintain that empty set is a subset of any sets.


*We want to maintain a theorem that says the union of open sets is an open set.
Now watch that the definition of an open set is the same as saying it is the union of all sets of all open balls in the set. As we want to maintain that empty set is a subset of any sets (reason number 1), we have that empty set is a subset of all sets of those open balls, thus empty set is a subset of an open set. As we also want to maintain the theorem mentioned in reason number 2 to be true, there is no other way than to define an empty set as an open set.
For the last part of the question of why the set $A$ and empty set are both also close is to remember that the definition of a close set is:
$$
\hbox{If $B \subset A$, then $A \setminus B$ is close $\iff B$ is open.}
$$
If we set $B = A$, then we have $\emptyset = A \setminus A$ in which by the definition above is close because $B = A$ is open. If we set $B = \emptyset$, then we have $A = A \setminus \emptyset$ which is close because $B = \emptyset$ is open.
A: Of course, the answer depends on how you define "open" and "closed" sets of $\mathbb R$. There are many equivalent definitions.
Here is a common one when we are considering $\mathbb R$ as a metric space with the absolute value norm: we say that $E\subset\mathbb R$ is open if every point of $E$ is an interior point of $E$; we say that $E$ is closed if every limit point of $E$ is a point of $E$.
According to these definitions, the assertion that $\varnothing$ is open is vacuously true. The assertion that $\mathbb R$ is open is true because every neighbourhood of every point of $\mathbb R$ is a subset of $\mathbb R$, and so it follows a fortiori that each point of $\mathbb R$ has a neighbourhood which is included in $\mathbb R$. Notice that neighbourhoods themselves are defined as certain subsets of $\mathbb R$, making this whole business rather trivial.
The assertion that $\varnothing$ is closed is also fairly easy to verify. If $x\in\mathbb R$ were a limit point of $\varnothing$, then every neighbourhood of $x$ would contain a point in $\varnothing$. Since the empty set is, well, empty, this is impossible. So the empty set has no limit points, and the implication is vacuously true. Since $\mathbb R$ is the metric space under consideration, for $x$ to be considered to be a limit point of $\mathbb R$, it has to member of $\mathbb R$, and so $\mathbb R$ is closed as well.
This proof can be shortened if we have previously demonstrated that a set is open if and only if its complement is closed.
