**Problem 1.**

… sequence $(x_n)$ in *S* there is a subsequence $(x_{n_k})$ which converges to some $x\in S$.

**Problem 2.**

… a sequence $(x_n)$ in *S* for which no subsequence converges to a point in *S*.

**Problem.3.**

A subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded (the Heine-Borel Theorem).

**Problem 4.**

If *S* is a single point *x*, there is only one sequence contained in *S*, the sequence *x*, *x*, … which converges itself.

**Problem 5.**

This fact is implied by the fact that the subsequence converges. In particular, if the subsequence is $(x_{n_k})$, then by definition of convergence for any $\epsilon>0$ every term with $k>N$ for some *N* is in the $\epsilon$-neighborhood of *x*. Since there are an infinite number of these, and they are all elements of the original sequence $(x_n)$, the statement must be true. $\blacksquare$

**Problem 6.**

We must show how to construct a subsequence $x_{n_1},x_{n_2},x_{n_3},\ldots$ which converges to *x*. Let $x_{n_1}$ be any element within the 1-neighborhood of *x* (that is, anything within distance 1 of the limit). Let $x_{n_2}$ be any element with $n_2>n_1$ (that is, some element of the sequence further out than $n_1$) and within the $\frac12$-neighborhood of *x*. Continue in this manner to select $x_{n_i}$ with $n_i>n_{i-1}$ in the $\frac1i$-neighborhood of *x*. Since there are an infinite number of points within each neighborhood of *x* to choose from, we can always find such points.

How can we prove that the subsequence $x_{n_k}$ converges to *x*? It suffices to show that every $\epsilon$-neighborhood of *x* contains some "tail" of the sequence, that is all $x_{n_k}$ with $k>N$ for some *N*. Choose *N* such that $0<\frac1N<\epsilon$. By construction if $k>N$ then $d(x_{n_k},x)<\frac1N<\epsilon$ since everything past $k=N$ is in a $\frac1{N+1}$-neighborhood or smaller. $\blacksquare$.