The Tychonoff Product Theorem

Show that the product of an arbitrary collection of Tychonoff spaces, with the product topology, also is Tychonoff.

Show that the product of an arbitrary collection of Hausdorff spaces, with the product topology, also is Hausdorff.

Show that the product topology on $n$-fold product of $\mathbb {R}$ is the same as the metric topology on $\mathbb {R}^{n}$ induced by the Euclidean metric.

Let ($X, \rho_{1}$) and ($Y, \rho_{2}$) be metric spaces. Show that the product topology on $X \times Y$, where $X$ and $X$ have the topologies induced by their respective metrics, is the same as the topology induced by the product metric

Show that if $X$ is a metric space with metric $\rho$, then

also is a metric on $X$ and it induces the same topology as the metric $\rho$.

Consider the countable collection of metric spaces $\left\{ \left(X_{n}, \rho_{n}\right)\right\}_{n = 1}^{\infty }$. For the Cartesian product of these sets $X = \prod_{n = 1}^{\infty } X_{n}$, define $\rho : X \times X \rightarrow \mathbb {R}$ by

Use the preceding problem to show that $\rho$ is a metric on $X = \prod_{n = 1}^{\infty } X_{n}$ which induces the product topology on $X$, where each $X_{n}$ has the topology induced by the metric $\rho_{n}$.

Consider the set $X = \mathbb {R}^{\mathbb {R}}$ with the product topology. Let $E$ be the subset of $X$ consisting of functions that take the value 0 on a countable set and elsewhere take the value 1. Let $f_{0}$ be the function that is identically zero. Then it is clear that $f_{0}$ is a point of closure of $E$. But there is no sequence $\left\{ f_{n}\right\}$ in $E$ that converges to $f_{0}$, since for any sequence $\left\{ f_{n}\right\}$ in $E$ there is some $x_{0} \in \mathbb {R}$ such that $f_{n}\left(x_{0}\right) = 1$ for all $n$ and so the sequence $\left\{ f_{n}\left(x_{0}\right)\right\}$ does not converge to $f_{0}\left(x_{0}\right)$. This shows, in particular, that $X = \mathbb {R}^{\mathbb {R}}$ is not first countable and therefore not metrizable.

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.

A product $I^{A}$ of unit intervals is called a (generalized) cube. Prove that every compact, Hausdorff space $X$ is homeomorphic to a closed subset of some cube. (Let $\mathcal{F}$ be the family of continuous real-valued functions on $X$ with values in $[0,1]$. Let $Q = \Pi_{f \in \mathcal{F}} I_{f}$. Then, since $X$ is normal and compact, the mapping $g$ of $X$ into $Q$ that takes $x$ into the point whose $f$-th coordinate is $f(x)$ is one-to-one, continuous, and has closed image.)

Let $Q = I^{A}$ be a cube, and let $f$ be a continuous real-valued function on $Q$. Then, given $\epsilon > 0$, there is a continuous real-valued function $g$ on $Q$ for which $\left|f-g\right| < \epsilon$ and $g$ is a function of only a finite number of coordinates.