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?
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).
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.
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.
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.