I can't seem to prove this: Let G be an equicontinuous subset of the continuous functions between E and F. Then the open point topology on G is jointly continuous.

For the definition of equicontinuity, here is a link: Question regarding uniform spaces and equicontinuity


Going by definitions found online, I assume you want to prove the following:

We have $E$ a topological space, $F$ a uniform space (in its corresponding topology). We have $G \subseteq C(E,F)$ equicontinuous (as in the linked definition). We consider $G \subseteq F^E$, where the last space has the product (or pointwise) topology. Then the mapping $A: G \times E \rightarrow F$, defined by $A(f, x) = f(x)$ is continuous.

Let $(f,x)$ be a fixed point in $G \times E$ and we want to show continuity of $A$ at this point. So pick a basic neighbourhood of $f(x)$ in $F$, and as this topology comes from a uniformity, we can take this neighbourhood of the form $W[f(x)]$, where $W$ is a member of the uniformity on $F$. Then find a symmetric uniformity member $V$ such that $V \circ V \subseteq W$. Now apply the definition of equicontinuity of $G$ at $x$ for this $W$, so we have an open neighbourhood $U$ of $x$, such that $$\forall y \in U: \forall g \in G: (g(y), g(x)) \in V\text{.}$$

Define the following open set $O$ for the pointwise topology on $F^E$:

$$O = \{h \in F^E \mid h(x) \in V[f(x)] \}$$

Clearly, $O \cap G$ is a basic neighbourhood of $f$ in the pointwise topology. Then I claim that $A[(O \cap G) \times U] \subseteq W[f(x)]$, which shows the continuity of $A$ at $(f,x)$.

Proof of the claim: let $g \in G \cap O$ and $y \in U$. Then $g(x) \in V[f(x)]$ because $g \in O$, and so $(f(x), g(x)) \in W$ and by the choice of $U$ and as $g \in G$, $(g(y), g(x)) \in V$. So $(f(x), g(y)) \in V \circ V \subseteq W$, so $g(y) \in W[f(x)]$ by definition.

  • $\begingroup$ Thank you Henno, solution well written and understood :) $\endgroup$ – User666x Jan 12 '15 at 21:44
  • $\begingroup$ @AvivEshed glad to be able to help! $\endgroup$ – Henno Brandsma Jan 12 '15 at 21:50

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