The product topology and compactness in $\reals[n]$

Suppose that $X$ and $Y$ are topological spaces and consider their product $X \times Y = \left\{ (x, y) : x \in X, y \in Y\right\}$. Define a set $W \subset X \times Y$ to be open if given any $(x, y) \in W$, then there are open sets $U$ in $X$ and $V$ in $Y$ so that $(x, y) \in U \times V \subset W$.

Verify that open sets as defined above satisfy the three properties required of a topology.

Proposition 1.5.4: Suppose $X$ is a metric space and $x \in X$. If $\left\{ s_n\right\}$ is a sequence so that for any ball about $x$, the ball contains an infinite number of the $s_n$ (this means that there are an infinite number of $n$ so that $s_n$ is in the ball), then there is a subsequence which converges to $x$.

The proof begins as follows: choose $n_1$ so that $s_{n_1}$ is contained in the ball of radius $1$ about $x$. Since there are an infinite number of the $s_n$ in the ball of radius $\tfrac {1}{2}$ about $x$, we can find $n_2$ so that $n_2 > n_1$ and $s_{n_2} \in B(x, \tfrac {1}{2})$. Inductively, we then use the same idea to choose $n_1 < n_2 < n_3 < \cdots$ so that $s_{n_j} \in B(x, 1/j)$. This will give a convergent subsequence.

Fill in the details in this proof: show that the resulting subsequence $\left\{ s_{n_j}\right\}$ does converge to $x$.

Show that if $\left\{ s_n : n \in \mathbb {N}\right\}$ is finite, then the sequence has a convergent subsequence.

Let $(X, d_X)$, $(Y, d_Y)$ be metric spaces. Then $f : X \to Y$ is said to be uniformly continuous if given $\epsilon > 0$ there exists a $\delta > 0$ such that for $x_1, x_2 \in X$, $d_X(x_1, x_2) < \delta$ implies $d_Y(f(x_1), f(x_2)) < \epsilon$.

Show that $f$ uniformly continuous implies $f$ is continuous, but construct an example to show that the converse does not hold.

Let $f : X \to Y$ be a continuous map of the compact metric space $(X, d_X)$ to the metric space $(Y, d_Y)$. Show that $f$ is uniformly continuous.