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The epsilon-delta definition of continuity:

A function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ is continuous at point $x_0 \in \mathbf{R}$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that whenever $|x–x_0| < \delta$ then $|f(x)–f(x_0)| < \epsilon$

Now from this, how shall we deduce that the converse statement is true?

Converse statement:

If a function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ is continuous at point $x_0 \in \mathbf{R}$, then for every $\epsilon > 0$ there exists a $\delta > 0$ such that whenever $|x–x_0| < \delta$ then $|f(x)–f(x_0)| < \epsilon$

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  • $\begingroup$ Those statements coincide. $\endgroup$
    – lulu
    Jul 8, 2019 at 15:10
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    $\begingroup$ You wrote the same thing twice. $\endgroup$ Jul 8, 2019 at 15:10
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    $\begingroup$ A def is an iff : a function $f$ is continuous at..." iff "for every $\epsilon > 0$ ..." There is nothing to prove : the def provides a criteria to be "assessed" : check if the defining condition is satisfied; if yes, you are entitled to assert that the function is continuous. $\endgroup$ Jul 8, 2019 at 15:13
  • $\begingroup$ Please have a more careful look at it: First statement says: "2 is true if 1 is true". The second statement says: "If 2 is true, then 1 is true" $\endgroup$
    – Joe
    Jul 8, 2019 at 15:14
  • $\begingroup$ Maybe Converse ? $\endgroup$ Jul 8, 2019 at 15:19

2 Answers 2

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Long comment

See T.Tao, Analysis, I, page 227 :

Definition 9.4.1 (Continuity). Let $X$ be a subset of $\mathbb R$, and let $f : X → \mathbb R$ be a function. Let $x_0$ be an element of X. We say that $f$ is continuous at $x_0$ iff we have :

$\lim_{x → x_0;x∈X} f(x) = f(x_0).$

Thus, there is nothing to prove.

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  • $\begingroup$ "The def is iff ". That was what I actually thought. But in a number of websites including wikipedia, it can't be found. Anyway thanks for the answer/comment. $\endgroup$
    – Joe
    Jul 8, 2019 at 15:36
  • $\begingroup$ @Joe - Usually, only the part "The function $f$ is continuous at some point $c$ of its domain if the limit of $f(x)$ ..." is used because what we need is the "defining criteria" to be verified in order to conclude that $f$ is continuous. $\endgroup$ Jul 8, 2019 at 17:03
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This is not the inverse statement. You need to change the quantifiers as well. That is if the original statement had an "for every" or "for all" the negation would have "there exists" and vice versa. Moreover if you have a statement of the form $A$ implies $B$ the negation would be $A$ and (the inverse of $B$). So for instance the negation of "$|x-x_0|<\delta\Rightarrow |f(x)-f(x_0)|<\varepsilon$" is "$|x-x_0|<\delta$ and $|f(x)-f(x_0)|\geq\varepsilon$".

In other words the negation of

"for every $\varepsilon>0$ there exists $\delta>0$ such that whenever $|x-x_0|<\delta$ then whenever $|x-x_0|<\delta$ we have $|f(x)-f(x_0)|<\varepsilon$"

would be

"There exists $\varepsilon>0$ such that for all $\delta>0$ there exists $x$ such that $|x-x_0|<\delta$ and $|f(x)-f(x_0)|\geq \varepsilon$."

Tip: It is the best to write the statement with the quantifiers $\forall$ and $\exists$ instead of words. That way it is much easier to replace each $\forall$ with $\exists$ and vice-versa.

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  • $\begingroup$ Sorry for not being careful in the first place. I actually meant "converse" of the statement. $\endgroup$
    – Joe
    Jul 8, 2019 at 15:23

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