Why is the empty set bounded? Why is the empty set bounded below and bounded above? If it has no elements, how can you say that an upper or lower bound exists?
 A: Alternatively, a set $S$ is bounded if there exist numbers $a$ and $b$ such that:
$$S\subseteq[a,b]$$
Now, $\varnothing$ is a subset of every set, so…
A: A set in a metric space is bounded if, and only if, there exists a ball (of finite radius) containing it. Then trivially the empty set is contained in a ball (actually in every ball).
A: Recall that implication has the property that when the assumption is false, the implication is true. In other words, if $P$ is false, then $P\implies Q$ is true.
Let $S$ be a set of real numbers. Then $M$ is an upper bound for $S$ if the following implication holds, $$s\in S\implies s \leq M.$$ Now let us examine the case for the empty set, $\emptyset$. 
Proposition Let $M$ be any real number. Then $M$ is an upper bound for the empty set (of real numbers).
Proof:
Since the statement, $s\in\emptyset$ is false, the implication, $$s\in\emptyset\implies s \leq M$$ is true.
Note: An almost identical proof works for the lower bound case.  A nice little slogan to remember here is 
All things are true about the MEMBERS of the empty set.
A: Argue by contradiction. Suppose $\emptyset$ is unbounded. Then for every $M > 0$ there is a point $x \in \emptyset$ such that $|x| > M$. But this contradicts that the empty set has no elements.
A: We can prove it by contradiction. Let S=∅ be the empty set and let S does not have any upper bound i.e. there exists no real number M such that M is greater than or equal to every x∈S, which implies there exist at least one element x∈S such that x>M, which is a contradiction since S is an empty set and has no elements.
Similarly if S does not have any lower bound then there exist no real number m which is smaller than or equal to every element of S, implies there exist at least one element x∈S such that x<m, again a contradiction since S is an empty set and has no elements.
Hence our assumption is wrong and S has both upper and lower bounds in fact every real number is an upper bound and every real number is a lower bound of an empty set S.
But an empty set has no least upper bound since we don't have any smallest real number, similarly an empty set has no greatest lower bound since there exists no greatest real number.
A: 
In mathematical analysis and related areas of mathematics, a set is
  called bounded, if it is, in a certain sense, of finite size.
  Conversely, a set which is not bounded is called unbounded. - Wikipedia definition.

Now there is only one "nothing" just like there is only one 1, or only one 2, or say only one zero. There may be 2 cats or 2 dogs or 2 giraffes but 2 is 2. there is only one 2 along the entire number-line. similarly there are only one "nothing". 
so the empty set, the content (nothing) is finite in number, which is 1. 
so empty set should be finite. 
please correct me i may make mistake im not math people. 
A: Note: this is an informal explanation and not a rigorous proof. It's just meant to strengthen your intuition (I hope!)
It might help to think of upper and lower bounds as limits on the size of a set's elements. Sets are either unbounded or bounded.
Let's look first at an unbounded set, $A$. If $A$ is unbounded, its elements get arbitrarily large. Large positive and "large negative". They have no upper limit to their absolute value. They "go off to infinity, and to negative infinity."
Bounded sets are sets that don't do this.
A set $S$ has an upper bound $b$ if no element of $S$ is greater than $b$. $S$ doesn't "go off to infinity". It doesn't even get past $b$
The empty set certainly doesn't "go off to infinity or negative infinity". It doesn't have elements greater than $b$, no matter what number $b$ is. It doesn't have elements less than $b$ either.
Every number is both an upper and lower bound of the empty set, a size restriction so impossibly restrictive that no real numbers can obey.
The empty set is so tightly bound up, so tightly restricted, it doesn't even have any elements!
A: We can say that the empty set is bounded if its not in R; that is if the empty set is the complement of R then we can say is bounded, because it lies in the extension real numbers.But if the empty set is a subset of R then it may be bounded or unbounded. If we assume it finite then its bounded set but if we look at its supremum and its infimum we see that they are not real numbers so the set is unbounded. I see that the empty set is both bounded and unbounded.
