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Let $B$ be a non-empty set equipped with the discrete topology, and let $A$ be an infinite set. Then $B^A$ is the set of all functions $f:A\to B$.

I have to verify some elementary properties of the product topology that $B^A$ inherits:

Prove that $U\subset B^A$ is open iff for all $f\in U$ there exists a finite $E\subset A$ such that all maps $g:A\to B$ with $g|_E=f|_E$ belong to $U$.

I don't know how to check this.. I know I should think of $B^A$ as $\prod_{a\in A}B_a$ where $B_a=B$ are copies of $B$ for all $a\in A$. The basis open sets are then $\prod_{a\in A}U_a$ where $U_a=B_a$ for all but finitely many $a$. Since $B$ has the discrete topology the $U_a$ which are not $B_a$ can be an arbitrary subset of $B$. Then the open sets are unions of the basis open sets.

But how does this give me the statement? Can someone provide any help?

Dr. Heinz Doofenshmirtz
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2 Answers2

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Let $\pi_a: B^A \to B$ be the map defined by $\pi_a(f)=f(a) \in B$. (these are just the projections if you look at it as a Cartesian product.) The product topology on $B^A$ is the smallest topology that makes all $\pi_a, a \in A$ continuous, which means (by standard facts about subbases and bases) that a base for its open subsets is given by all finite intersections of the form $\pi_a^{-1}[\{b\}]$, with $a \in A$ and $b \in B$ (as all $\{b\}$, $b \in B$ form a base for the discrete topology on $B$). So a basic open subset is of the form

$$[a_1, a_2, \ldots, a_n; b_1, b_2, \ldots b_n] := \bigcap_{i=1}^n \pi_{a_i}^{-1}[\{b_i\}] = \{f: A \to B: \forall i \in \{1, \ldots n\}: f(a_i)= b_i\}$$

where, $n \in \mathbb{N}$ and the $a_i$ and $b_i$ are elements of $A$ resp. $B$.

Now if $U$ is open and $f \in U$, we thus must have some $[a_1, a_2, \ldots, a_n; b_1, b_2, \ldots b_n]$ such that $$f \in [a_1, a_2, \ldots, a_n; b_1, b_2, \ldots b_n] \subseteq U$$

by the fact that $U$ is a union of basic open sets. And now if $g=f$ on the finite set $E:= \{a_1,\ldots,a_n\}$ it means exactly that $g \in [a_1, a_2, \ldots, a_n; b_1, b_2, \ldots b_n]$ too and so in $U$ as well.

The reverse is quite similar: if $U$ is some set and for each $f$ we can find such a finite subset $E$ then, if $E=\{a_1,\ldots, a_m\}$, we have $f \in U(f):= [a_1, a_2, \ldots a_m; f(a_1), \ldots f(a_m)]$ and the condition ensures that if $g \in U(f)$ then $g=f$ on $E$ and so $g \in U$, hence $f \in U(f) \subseteq U$ and so $U$ is open. (All $f$ in it are interior points).

Henno Brandsma
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1

Let $U\subseteq B^A$ be open, and $f\in U$.
Then there exists a basis open set $V\subseteq U$ with $f\in V$.
Now let $E$ be the finite set of indices $i\in A$ such that $\pi_i(V)\ne B$, where $\pi_i:B^A\to B$ is the $i$th projection (i.e., evaluation at $i$).

If $g|_E=f|_E$, then $\pi_i(V)=B$ for all $i\notin E$ implies $g\in V$.

Conversely, if for every $f\in U$, there's such a finite $E\subset A$, then the set $V_f:=\{g\in B^A: g|E=f|E\}$ is a basis open set, and $f\in V_f\subseteq U$, so $U=\bigcup_{f\in U} V_f$ is open.

Berci
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