Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, and let $Y^X : = \{f| f: X\to Y\}$ be the space of all functions from $X$ to $Y$. If $x_0 \in X$ and $V$ is an open set in $Y$, let $V^{(x_0)} \subseteq Y^X$ be the set
$$ V^{(x_0)} : = \{f \in Y^X: f(x_0) \in V\}.$$
If $E$ is a subset of $Y^X$, we say that $E$ is open if for every $f \in E$, there exists a finite number of points $x_1, … , x_n \in X$ and open sets $V_1, …, V_n \subseteq Y$ such that
$$f \in V_1^{(x_1)} \cap \ldots \cap V_n^{(x_n)} \subseteq E.$$
1.Show that if $F$ is the collection of open sets in $Y^X$, then $(Y^X, F)$ is a topological space.
2.For each natural number $n$, let $f^{(n)} : X \to Y$ be a function from $X$ to $Y$, and let $f: X \to Y$ be another function from $X$ to $Y$. Show that $f^{(n)}$ converges to $f$ in the topology $F$ if and only if $f^{(n)}$ converges to $f$ pointwise.
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I can show that $\bigcap_{i=1}^n E_i$ and $\bigcup_{i\in I} E_i$ are open sets (for $I$ possibly infinite). $\emptyset \in F$ trivially. However, I don't know how to show that $Y^X \in F$.
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For every $E \in F$ containing $f$, there exists $N$ such that for $n \ge N$, $f_n \in E$ ($f_n$ converges to $f$ in a topological space). For every $x \in X$ and $\epsilon >0$, there exists $N$ such that for $n \ge N$, $d_Y(f_n(x), f(x)) < \epsilon$ ($f_n$ converges pointwise to $f$). I am not sure how to connect these two definitions.
Any help would be greatly appreciated.
Best Answer
The fact that $F$ is a topology can be shown rather easily: for any $f \in Y^X$ we can use $Y^{(x)}$ for any point $x \in X$ ($X$ is non-empty, I assume): $Y$ is open and $Y^{(x)}$ is just equal to $Y^X$. $\emptyset$ is open by void truth.
If $O_i, i \in I$ is a collection of sets in $F$, then so is $O:=\bigcup_i O_i$: let $f \in O$ be arbitrary and find $i_0 \in I$ such that $f \in O_{i_0}$.
As $O_{i_0} \in F$ there exist finitely many $V_1, \ldots, V_n$, open in $T$ and finitely many points $x_1, \ldots x_n \in X$ such that $$f \in \bigcap_{i=1}^n V_i^{(x_i)} \subseteq O_{i_0} \subseteq O$$
which shows that $O$ also obeys the condition to be in $F$ (just by virtue of being a superset, really).
For the finite intersection case, it suffices to consider the case of two sets in $F$, say $U,V \in F$ and let $f \in U \cap V$. In particular $f \in U$ and $U$ open implies that there are finitely many open sets $U_1, \ldots, U_n$ and as many points $x_1, \ldots, x_n \in X$ such that
$$f \in \bigcap_{i=1}^n U_i^{(x_i)} \subseteq U$$
and $V$ open means that there are also finitely many $V_1, \ldots, V_m$ and as many $x'_1, \ldots, x'_m \in X$ such that
$$f \in \bigcap_{i=1}^m V_i^{(x'_i)} \subseteq V$$
Now, for the points that occur both as some $x_i$ and some $x'_j$ (so in the intersection of the two finite sets) we use $U_i \cap V_j$ as the open set for that point, and for the other "unique" points we just use its given $U_i$ or $V_i$ and then the finite intersection of all these (but still finitely many) sets of the for $O^{(x)}$ still contains $f$ and is now contained in $U \cap V$ because both sets of demands on $f$ are fulfilled. So the only subtlety is that we need an intersection for the common points.
As for (ii): if $f^{(n)} \to f$ then for any $\varepsilon>0$ and every $x \in X$, we can use $B(f(x), \varepsilon)^{(x)}$ (which is open, check that!) as an open neighbourhood for $f$ and see that $N$ exists so that
$$\forall n \ge N: f^{(n)} \in B(f(x), \varepsilon)^{(x)}$$
which just says
$$\forall n \ge N: f^{(n)}(x) \in B(f(x), \varepsilon)$$ or equivalently
$$\forall n \ge N: d(f(x), f^{(n)}(x)) < \varepsilon $$
so in total we showed:
$$\forall \varepsilon>0: \forall x \in X: \exists N: \forall n \ge N: d(f(x), f^{(n)}(x)) < \varepsilon $$
which is pointwise convergence.
The reverse is not hard, namely that a pointwise convergent sequence converges in $(Y^X,F)$, using that open balls form a base for $Y$ and peeling off the definitions as before.