1
$\begingroup$

Let $(E, d)$ be a metric space and $\mathcal F$ a collection of real-valued functions on $E$. Assume that for all $x,x_n \in E$ with $n\in \mathbb N$, $$ x_n \to x \iff [f(x_n) \to f(x) \quad \forall f \in \mathcal F]. \quad (\star) $$

Let $\tau$ be the metric topology on $E$ induced by $d$ and $\tau'$ the initial topology induced by $\mathcal F$. Clearly, $\tau' \subset \tau$. It's possible that $\tau' \neq \tau$.

My question Are there some conditions on $\mathcal F$ that ensures $\tau' = \tau$?

Thank you so much for your elaboration!

$\endgroup$
0

1 Answer 1

2
$\begingroup$

If $\mathcal{F}$ is countable, $\tau'$ is metrizable with a compatible metric $\rho$ given by $$\rho(x,y)=\sum_n 2^{-n}~|f_n(x)-f_n(y)|\wedge1$$ for some enumeration of $\mathcal{F}$.

Since the topology of a metric space is determined by the convergent sequences, this is a sufficient condition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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