Properties of Product Topologies

Munkres Topology

Theorem 1) Let $f:A\to {\prod }_{\alpha \in J}{X}_{\alpha }$ be given by the equation

$$f(a)=(f_{\alpha }(a){)}_{\alpha \in J}$$

where $f_{\alpha }:A\to X_{\alpha }$ for each $\alpha \in J$. If we give ${\prod }_{\alpha \in J}{X}_{\alpha }$ the product topology, $f$ is continuous if and only if each function $f_{\alpha }$ is continuous.

Proof) The forward direction is quite trivial. For any $\beta \in J$, we can define the projection function ${\pi }_{\beta }:{\prod }_{\alpha \in J}{X}_{\alpha }\to X_{\beta }$ such that $f_{\beta }={\pi }_{\beta }\circ f$. Clearly, ${\pi }_{\beta }$ is continuous by how the product topology is defined; as $f_{\beta }$ is the composition of two continuous functions, it is continuous for each $\beta \in J$. Note this argument also works in the box topology, as it is finer than the product topology; the reverse direction, however, doesn’t necessarily hold in the box topology.

To prove the reverse, it suffices to show that if each $f_{\alpha }$ is continuous, the inverse image of each subbasis element in ${\prod }_{\alpha \in J}{X}_{\alpha }$ is open in $A$. Note that we can represent each subbasis element as ${\pi }_{\beta }^{-1}(U_{\beta })$ for some $\beta \in J$ and open set $U_{\beta }$ in $X_{\beta }$. Then,

$$f^{-1}({\pi }_{\beta }^{-1}(U_{\beta }))=f_{\beta }^{-1}(U_{\beta })$$

which is open by the continuity of $f_{\beta }$.

Theorem 2) Let $U_i$ be a countable collection of metrizable topological spaces, where $d_i$ is a metric that induces the topology of $U_i$. The cartesian product $\prod U_i$ under the product topology is metrizable.

Proof) Let ${\overline{d}}_i=\min \{d_i,1\}$ and define the metric

$$D(x,y)=\sup \{\frac{{\overline{d}}_i(x_i,y_i)}{i}\}$$

in $\prod U_i$. We claim the topology induced by $D$ is precisely the product topology in $\prod U_i$. First, take an arbitrary basis element $\prod X_i$ in $\prod U_i$, where $X_i=U_i$ for all but finitely many $i$, in which $X_i$ is an open proper subset of $U_i$. Let $i_1,i_2,\dots ,i_n$ be the indices for which $X_i\ne U_i$, and take an arbitrary point $x$ in this basis element. As $d_i$ induces the topology of $U_i$ and as $X_i$ is open in $U_i$, for each $i_j$ there exists some $k_j>0$ such that the $k_j$-neighborhood of $x$ in the ${\overline{d}}_{i_j}$ metric is a subset of $U_{i_j}$. Taking $k=\min (k_j)>0$, we can see the $k$-neighborhood of $x$ in the $D$ metric of $\prod U_i$ is a subset of the basis element $\prod X_i$. Therefore, the $D$ metric topology is finer than the product topology $\prod U_i$.

Conversely, take an arbitrary basis element of the $D$-metric topology by choosing a point $x\in \prod U_i$ and $r>0$ and taking the $r$-neighborhood $W\subset \prod U_i$ of $x$. There must exist some $n$ such that $nr\ge 1$ and $(n-1)r<1$; a point $y$ is in $W$ iff $\sup \{{\overline{d}}_i(x_i,y_i)/i\}<r$, which occurs iff $d_i(x_i,y_i)<ir$ for $i<n$ and $y_i\in U_i$ for $i\ge n$. That is, $W$ is precisely the set

$$W=\prod _{i=1}^{r-1}{N}_{d_i,ir}(x_i)\prod _{i=r}^{\infty }{U}_i$$

which is clearly open in the product topology. Thus, the product topology is finer than the $D$ metric topology, implying the two are identical.

Theorem 3) Let $(U_{\alpha }{)}_{\alpha \in J}$ be an arbitrary collection of connected spaces. The cartesian product $\prod U_{\alpha }$ in the product topology is also connected.

Proof) Note that finite products of connected spaces must be connected. Given a finite subset $W\subset J$, let $S_W$ be the subset of $\prod U_{\alpha }$ such that

$$S_W=\prod _{w\in W}{U}_w\prod _{w\notin W}\{0{\}}_{U_w}$$

S_W$isthenhomeomorphictothefiniteproduct$\prod_{w \in W} U_w$,whichisconnected,implying$S_W$isconnected.Let$Y$betheunionofall$S_W$swhere$W$isafinitesubsetof$J$.Asall$S_W$scontainthepoint$(0,0,\ldots)$,$Y$mustbeconnected.Weclaimthat$Y$isdensein$\prod U_\alpha$,whichforces$\prod U_\alpha = \overline{Y}$ to be connected as well.

Take any basis element $\prod X_{\alpha }$ in $\prod U_{\alpha }$, where $X_{\alpha }=U_{\alpha }$ except for finitely many indices $\alpha$. If we, for the sake of simplicity, say these indices are precisely $X_1,X_2,\dots ,X_n$, then the point $(x_1,x_2,\dots ,x_n,0,0,\dots )$ where $x_i$ is an arbitrary element of $X_i$ must be in $Y$. Therefore, $\prod X_{\alpha }\cap Y\ne \varnothing$, meaning any arbitrary open set in the product topology $\prod U_{\alpha }$ intersects with $Y$. Thus, $Y$ is dense in $\prod U_{\alpha }$ and we are done.