Originally Posted by

**kalagota** no, just at least one..

Let us consider the set $\displaystyle I$. Since it is bounded, there is an $\displaystyle L$ such that $\displaystyle I = [-L,L]$.

Bisect $\displaystyle I$ into $\displaystyle {I_1}' = [-L,0]$ and $\displaystyle {I_1}'' = [0,L]$.

At least one of these has the properties:

1. intersection with $\displaystyle I$ is not empty; and (WHY?)

2. Infinite (WHY?)

Name that set $\displaystyle I_1$. Again, bisect $\displaystyle I_1$ into $\displaystyle {I_2}'$ and $\displaystyle {I_2}''$. again, at least one of these has the properties:

1. intersection with $\displaystyle I_1$ is not empty; and (WHY?)

2. Infinite (WHY?)

Name that set $\displaystyle I_2$..

Do this continuously and form the sequence of nested intervals $\displaystyle \{I_n\}$.

Use the Property of Nested interval, $\displaystyle \exists \ x$ such that $\displaystyle x \in I_n$ for all $\displaystyle n$.

... now, can you do the rest? you only need to show that every neighborhood of $\displaystyle x$ contains infinitely many points from $\displaystyle I_n$ and hence from $\displaystyle I$. (maybe, two steps or three more.. Ü)