Consider the continuous function $f: \mathbb{R} \backslash \{0\} \rightarrow \mathbb{R}$ by $f(x)=1/x$. The graph of f is the hyperbola $xy=1$. One may conclude from the last exercise that $\mathbb{R} \backslash \{0\}$ is homeomorphic to $xy=1$.

Consider the metric spaces $(X_i,d_i),i=1 ,\cdots ,m$. Check that $d(x,y)=max_id_i(x_i,y_i)$ defined on $X= {\Pi ^m}_{i=1} X_i$ is a metric. The metric topology coincides with the product topology on $X$. This may be concluded from the fact that for $x=(x_1, \cdots ,x_m),\mathbb{B_d}(x,r) ={\Pi ^m}_{i=1} \mathbb{B_{d_i}}(x_i,r)$.

Show that projections $\pi_j:{\Pi^m}_{i=1}X_i \rightarrow X_j$ is continuous.

Verify that $\mathbb{B_b}$ and $\mathbb{B_p}$ are basis for a topology on ${\Pi^m}_\alpha X_\alpha$:

(1)  $\mathbb{B_b}=\{\Pi_{\alpha \in I } U_\alpha :U_\alpha \; is \; open \; in X_\alpha \}$ .

(2) $\mathbb{B_b}=\{\Pi_{\alpha \in I } U_\alpha \in \mathbb{B_b}:U_\alpha=X_\alpha$ for finitely many values of $\alpha\}$.

Suppose that each $X_\alpha$ contains a non-empty proper open subset $U_\alpha$. Show that $\Omega_\mathbb{B_p} = \Omega_\mathbb{B_b}$ iff $I$ has finite cardinality.

Show that the product topology is the smallest topology which makes all projections $\pi_\alpha$ continuous.

Let $f:A \rightarrow \Pi_\alpha X_\alpha$ be given by $f(a)=(f_\alpha(a))$, where the functions $f_\alpha : A \rightarrow X_\alpha$ are given. Suppose $\Pi_\alpha X_\alpha$ has either box topology or product topology. Show that if f is continuous then each $f_\alpha$ is continuous.

Show that $D : \mathbb{R^ \omega} \times \mathbb{R^ \omega} \rightarrow \mathbb{R}$ given by

$D((x_n),(y_n)):=sup_n \frac{min \{|x_n-y_n|,1\}}{n}$

defines a metric on $\mathbb{R^ \omega}$ . Verify further that $D \le d_u$ , where $d_u$ is the uniform metric.

Show that any finite set is compact.

Let $K \subseteq Y \subseteq X$. Show that $K$ is compact in $X$ iff $K$ is compact in $Y$.

We invoke a classical result from Analysis, which characterizes all compact subsets of $\mathbb{R^n}$ .

Let p be a polynomial in the real variables $x,y$. Then the zero set $Z(p)$ of $p$ is always closed. However, $Z(p)$ may or may not be compact. For instance, if $p(x,y)=x^2=y^2-1$then $Z(p)$ being the unit circle is compact. If $p(x,y)=x$ then $Z(p)$ is the Y -axis, which is certainly non-compact.