Tychonoff's Theorem

Prove the stated assertion. For the product $X \times Y$ of metric spaces $X$ and $Y$ (see problem 17) to be a compact set, it is necessary and sufficient that the spaces $X$ and $Y$ be compact sets.

Let $\left\{ M_{i}\right\}_{i=1}^{n}$ be compact metric spaces. Prove that $M:=M_{1} \times M_{2} \times \cdots \times M_{n}$ is a compact metric space with respect to metric $d$ as in Exercise 2.1.14.

For the countable Cartesian product of metric spaces considered in Proposition 7, show that the product is compact if and only if each factor space is compact.

Proposition 7: Let $X = \prod_{n=1}^{\infty } X_{n}$ be the Cartesian product of the countable collection of metric spaces $\left\{ (X_{n}, \rho_{n})\right\}_{n=1}^{\infty }$. For points $u = \left\{ u_{n}\right\}$ and $v = \left\{ v_{n}\right\}$ in $X$, define $\rho (u, v) = \sum_{n=1}^{\infty } \frac{1}{2^{n}} \cdot \frac{\rho_{n}(u_{n}, v_{n})}{1 + \rho_{n}(u_{n}, v_{n})}$. Then $\rho$ is a metric on $X$ for which sequential convergence is componentwise convergence in each component metric space $(X_{n}, \rho_{n})$.

Let $X$ denote the discrete topological space with two elements. Show that $X^{N}$ is homeomorphic to the Cantor set.

Using the Tychonoff Product Theorem and the compactness of each closed, bounded interval of real numbers prove that any closed, bounded subset of $\mathbb {R}^{n}$ is compact.

Provide a direct proof of the assertion that if $X$ is compact and $I$ is a closed, bounded interval, then $X \times I$ is compact.

Prove that the product of a countable collection of sequentially compact topological spaces is sequentially compact.

Let $\Omega = X_{i \in \mathbb {N}} X_{i}$ be a product of metric spaces. Define $\Omega_{n} = X_{i \leq n} X_{i}$ and $S_{n} = \mathrm{X}_{i \geq n} X_{i}$. For each $n$ let $K_{n}$ be a compact subset of $\Omega_{n}$, with the property that $H_{n}: = \cap_{i \leq n}\left(K_{i} \times S_{i+1}\right) \neq \emptyset$ for each finite $n$. Write $\pi_{i}$ for the projection map from $\Omega$ onto $\Omega_{i}$. Show that $H: = \cap_{i \in \mathbb {N}} H_{i}$ is a nonempty compact subset of $\Omega$, by following these steps. (Remember, for metric spaces, compactness is equivalent to the property that each sequence has a convergent subsequence.)