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I'm having a problem understanding the concept of continuity in metric spaces. I know that by definition $f$ is continuous in the metric sense if

for all $x \in X$ and $\epsilon > 0$ there exists $\delta > 0$ such that $x_1 \in X$ and $d_X(x_1, x) < \delta$ implies $d_Y (f(x_1), f(x)) < \epsilon$, where $(X, d_X)$ and $(Y, d_Y)$ are metric spaces and $f : X \rightarrow Y$ is a function.

Can someone please explain what this intuitively means?

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  • $\begingroup$ Continuous functions have no breaks. Consider a step function like $f(x)=0$ for $x\le 0$ and $f(x)=1$ for $x\gt 0$. Use the definition to see why this is not continuous at $x=0$ and you will get it. $\endgroup$ – John Douma Oct 21 '16 at 2:30
  • $\begingroup$ Intuitively, continuity specifies that we can safely wiggle around a bit in the domain without having crazy things happen. For each $\epsilon$ neighborhood in the image, we are guaranteed wiggle room of $\delta$ in the domain. $\endgroup$ – user332239 Oct 21 '16 at 2:45
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If you have a neighborhood, N(x), of points in X close enough to a point, x, in the domain (i.e. $\lt\delta$), there is a neighborhood of radius $\epsilon$ about f(x) that is the image of N(x). There is a $\delta$ for every $\epsilon \gt 0$.

In other words, if you get close to x in the domain, f maps these points to points in the range of f, which are close to f(x).

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