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I seem to understand topology more than analysis and was wondering if these definitions of continuity, which to me have more a topological flavor, are correct.


Suppose $X$ and $Y$ are metric spaces and let $f: X \to Y$.

$f$ is pointwise continuous on $X$ if:

for all $x \in X$ and for all $\epsilon>0$, there exists $\delta>0$ such that $B_\delta (x) \subset f^{-1}(B_\epsilon(f(x)))$.

$f$ is uniformly continuous on $X$ if:

for all $\epsilon>0$, there exists $\delta>0$ such that for all $x \in X$, $B_\delta (x) \subset f^{-1}(B_\epsilon(f(x)))$.

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    $\begingroup$ Yes. Looks fine to me! $\endgroup$ – Dzoooks Jun 27 '18 at 23:55
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Yes ,I would agree they are correct; in essence they differ by the placement of the $\forall x$ quantifier, you could say.

The first can be made really topological because the $\delta$ varies by point, and so is just another way of saying "some neighbourhood of $x$". If we denote by $\mathcal{N}_X(x)$ the set of all neighbourhoods of $x$ in a space $X$, the regular continuity can be formulated as

$$ \forall N \in \mathcal{N}_Y(f(x)) \exists N' \in \mathcal{N}_X(x): f[N'] \subseteq N$$ where the last inclusion could also have been written $N' \subseteq f^{-1}[N]$, to stay closer to your formulation.

The second one is not topological, but a notion that belongs to so-called uniform spaces, spaces with a uniform structure. If I choose the "entourages"-view so that $(X,d)$ has a metric uniform structure $\mathcal{E}_d$ generated by entourages of the form $\{(x,y): d(x,y) < \varepsilon\}$ etc. we formulate the uniform continuity as

$$\forall U \in \mathcal{E}_{(Y,d)}: (f \times f)^{-1}[U] \in \mathcal{E}_{(X,d)}$$ which looks like normal continuity between topological spaces (inverse image of a "special" subset in the co-domain is "special" in the domain).

The notion is not purely topological because the global "narrowness" $\delta$ cannot be really stated as a condition on open sets, but needs the metric here.

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