Originally Posted by

**kathrynmath** Ok, so i did this proof and I'm unsure about it?

I wnat to prove that $\displaystyle {a_n}$ has a convergent subsequence.

Proof:

Since $\displaystyle {a_n}$ is a bounded sequence of real numbers, it must have at least one accumulation point. Let A be that accum pt. By definition, A is an accum pt of $\displaystyle {a_n}$ iff each neighborhood contains points of $\displaystyle {a_n}$ which are not A. Let $\displaystyle \varepsilon>0$ be arbitrary. So, we can let $\displaystyle a_n$$\displaystyle _k$ elemnet of (A-$\displaystyle \varepsilon$,A+$\displaystyle \varepsilon$). As $\displaystyle \varepsilon--->0$,(A-$\displaystyle \varepsilon$,A+$\displaystyle \varepsilon$) accumulates at A. Therefore, A is an accum pt of $\displaystyle a_n$$\displaystyle _k$.

I'm afraid I wasn't specific enough and should have included the case for finitely many distinct terms.