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The following two are definitions of continuity of a function:

1) The function is continuous at every point $c$ in the domain. The function is continuous at a point $c$ if for any given $\epsilon > 0$, we can find a $\delta$ such that $\forall x \in (c-\delta, c+\delta)$ $ |f(x) - f(c)| < \delta$.

2) Inverse image of every open set is open.

Now, I have gone through the proof that these two are equivalent from the Rudin's book, but, how to understand this equivalence word by word intuitively. Which words in $(1)$ correspond to which word in $(2)$.

Martin Sleziak
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user96000
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    Equivalence of definitions does not imply a bijection between words of the formulation. – Hagen von Eitzen Oct 13 '13 at 13:27
  • In your last inequality in (1) it must $;\epsilon;$ on the right side – DonAntonio Oct 13 '13 at 13:49
  • This doesn't directly address your question but perhaps you are confused by why the topology affects continuity, i.e. how it can be that the pre-image of open sets being open gives you continuity. The following post might be helpful: http://math.stackexchange.com/questions/523198/what-does-it-mean-to-induce-a-topology/523235#523235 – snar Oct 13 '13 at 14:06

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First, notice that any open set in $\mathbb{R}$ is a union of open intervals. Thus, we will only consider intervals. Given an $\epsilon$ and $\delta$, you've got that $(a-\delta,a+\delta)$ and $\left(f(a)-\epsilon,f(a)+\epsilon\right)$ are open intervals containing $a$ and $f(a)$. Your definition $(1)$ then can be seen to be equivalent to the fact that given an open interval around $f(a)$, there is an open interval around $a$, call it $I$, small enough so that $f(I)\subseteq (f(a)-\epsilon,f(a)+\epsilon)$. This $I$ is your $(a-\delta,a+\delta)$. But, this says that $f^{-1}\left(f(a)-\epsilon,f(a)+\epsilon\right)$ is a union of open intervals, hence open.

J126
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