Is it already in literature this generalized notion of uniform continuity in an arbitrary topological space (not necessarily in exactly the same form)? Let $(X, T_{1})$, $(Y, T_{2})$ be topological spaces; let $f: X \to Y$. Then $f$ is called uniformly continuous-($T_{1}, T_{2}$) if and only if for every $U \in T_{2}$ there is some $V \in T_{1}$ such that $x,y \in V$ imply $f(x), f(y) \in U$.

In particular, if the codomain is restricted to be $\mathbb{R}$ with the standard topology, is the corresponding definition existing in literature?


  • 1
    $\begingroup$ This is almost the definition in term of uniformities ( entourage form). $\endgroup$ – Henno Brandsma Dec 9 '18 at 16:31
  • 2
    $\begingroup$ Look up uniform spaces. $\endgroup$ – Lee Mosher Dec 9 '18 at 16:50

The definition you've proposed is a weaker (instead of stronger) form of classical continuity. Indeed, if $f$ is continuous then for an open $U$ put $V:=f^{-1}(U)$ to satisfy your condition.

Now not every function that satisfies your property is continuous. Take

$$f:\mathbb{R}\to\mathbb{R}$$ $$f(x)=\begin{cases}x &\text{if }x \neq 0 \\ 1 &\text{otherwise} \end{cases}$$

With the standard topology on $\mathbb{R}$ on both sides we have that $f$ satisfies your condition but is not continuous. The point is that we can always choose $V$ to avoid discontinuity at $0$.

There's no way to define uniform continuity on a general topological space. For that you need some additional structure.


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.