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

for all x∈Xx \in X and ϵ>0\epsilon > 0 there exists δ>0\delta > 0 such that

x1∈Xx_1 \in X and dX(x1,x)<δd_X(x_1, x) < \delta implies dY(f(x1),f(x))<ϵd_Y (f(x_1), f(x)) < \epsilon,
where (X,dX)(X, d_X) and (Y,dY)(Y, d_Y) are metric spaces and f:X→Yf : X \rightarrow Y is a function.
Can someone please explain what this intuitively means?
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Continuous functions have no breaks. Consider a step function like f(x)=0f(x)=0 for x≤0x\le 0 and f(x)=1f(x)=1 for x>0x\gt 0. Use the definition to see why this is not continuous at x=0x=0 and you will get it.

– John Douma

Oct 21 at 2:30

Intuitively, continuity specifies that we can safely wiggle around a bit in the domain without having crazy things happen. For each ϵ\epsilon neighborhood in the image, we are guaranteed wiggle room of δ\delta in the domain.

– user332239

Oct 21 at 2:45

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1 Answer

1

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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. <δ\lt\delta), there is a neighborhood of radius ϵ\epsilon about f(x) that is the image of N(x). There is a δ\delta for every ϵ>0\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).