In metric spaces it is a well known fact that uniformly continuous functions are indeed continuous at any point.

What about uniform spaces? How can I prove this?
(with the definition of topology in terms of uniform structure from here)

  • $\begingroup$ what did you try? $\endgroup$ – Ittay Weiss Jun 28 '14 at 8:11
  • $\begingroup$ I'm trying to prove that $f^{-1}(V[f(x)])=(f^{-1}V)[x]$ for $V\in\Phi$ but it seems hopelessly or even wrong... What do you think? $\endgroup$ – C-Star-W-Star Jun 28 '14 at 8:17

Let $f: (X,\mathcal{U}) \rightarrow (Y, \mathcal{V})$ be uniformly continuous.

Let $V[f(x)]$ be a basic neighbourhood of $f(x)$ in $Y$. As $V \in \mathcal{V}$ is an entourage, and $f$ is uniformly continuous, $(f \times f)^{-1}[V] \in \mathcal{U}$. Call this last entourage $U$. Then $f[U[x]] \subset V[f(x)]$: $y \in f[U[x]] \rightarrow \exists x' \in U[x]: f(x') = y$, and so $(x', x) \in U$, which means that $(f \times f)(x',x) = (f(x'), f(x)) = (y, f(x)) \in V$, so $y \in V[f(x)]$. So $f$ is continuous at $x$.


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