0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.