I have some questions on logical implications of the definition of continuity. Here is the context:

  1. Let $(\Bbb R,d)$ a metric space with the standard metric.
  2. Define $\,\,f:\Bbb R \rightarrow\Bbb R\,$ to be a function, and let $x_0\in \Bbb R$
  3. $f$ is said to be continuous at $x_0$ $\iff [\forall\epsilon>0,\,\exists\delta>0;\,\forall x\in \Bbb R,\,|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon].$

My question is: Are the following implications true? I feel like I am mixing up some stuff.

1) $f$ is continuous at $x_0$ $\iff [\forall\epsilon>0,\,\exists\delta>0;\,\forall x\in \Bbb R,\,|x-x_0|<\delta\iff |f(\delta)|\le\epsilon].$ (So that it is sufficient to construct a $\delta$ satsifying $|f(\delta)|\le\epsilon).$

2) $f$ is continuous at $x_0$ $\iff|f(x-x_0)|\le f(x)-f(x_0)$ (I have a feeling I am mis-using the topological definition of continuity.

PARTIAL END: NO NEED TO READ BEYOND (but if you are feeling inspired, then please continue :)


If 1) and 2) are true, then to prove continuity, can we do something along the lines of the following:? (Please forgive the complete lack of rigor, I am just trying to run things on a probably false sense of intuition)

  • Fix $\epsilon=\epsilon_0>0$

  • We wish to find a $\delta>0$ such that $\forall x\in \Bbb R,\,|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon_0.$

  • But, there exists $g(x)\in \Bbb R:g(x)*|x-x_0|=|f(x)-f(x_0)|$

  • But $g(x)*|x-x_0|=|f(x)-f(x_0)|<g(x)*\delta$

  • And $g(x)*|x-x_0|=|f(x)-f(x_0)|<\epsilon_0$

  • So we wish to find a $\delta>0$ such that $\delta\le \frac {\epsilon} {g(x)}$

If my last part gives no insight however wrong it may be, please tell me.

  • 2
    $\begingroup$ Your property 1 cannot be correct: note that the condition $|f(\delta)\leq \epsilon$ does not deepnd on $x$ or on $x_0$; so you would be saying that provided that $f$ is “small” near zero, that means that all real numbers are close to $x_0$. That’s patently false. E.g., if $f(x)=x$, then it is certainly false that $|x-x_0|\lt \delta$ holds if and only if $|f(\delta)|\leq epsilon$; the latter holds for $\delta<\epsilon$, the former only if $x$ is $\delta$-close to $x_0$. $\endgroup$ – Arturo Magidin Apr 8 '18 at 18:27
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    $\begingroup$ Your property 2 cannot be correct either: $f$ would never be continuous at an absolute maximum, since $f(x)-f(x_0)$ would always be negative. $\endgroup$ – Arturo Magidin Apr 8 '18 at 18:29
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    $\begingroup$ I gave you an example of a function that is continuous everywhere, but fails to meet your condition when $\epsilon<10$: $f(x)=x^2+10$. Run through that specific example to see why it is just not true that $f(\delta)$ is bounded by $\epsilon$ even though you can make the images of all points close enough to $x_0$ end up close enough to $f(x_0)$. $\endgroup$ – Arturo Magidin Apr 8 '18 at 18:48
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    $\begingroup$ No, it still doesn’t work: First, you have two biconditionals; but biconditionals are not associative, so your statement is ambiguous. Second: your final condition does not even mention $x$. You are claiming that provided that $f$ Takes a small value on $x_0-\delta$ or on $x_0+\delta$, this implies that every real number is close to $x_0$. That’s just nonsense. Also, what the function does at $x_0+\delta$ and at $x_0-\delta$ does NOT tell you what it does in the interval $(x_0-\delta,x_0+\delta)$. (Cont) $\endgroup$ – Arturo Magidin Apr 8 '18 at 20:40
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    $\begingroup$ In the end, you seem to think that functions are linear: that if $f(u)$ is small, then $f(x+u)$ will be close to $f(x)$. You can conclude that when $f(x+u) = f(x)+f(u)$ (the function is additive); but most functions do not satisfy this condition. The fact that $f$ is small at a few points does not tell you anything about what $f$ does near $x_0$. $\endgroup$ – Arturo Magidin Apr 8 '18 at 20:55

You are mixing things up.

The statement

$$ \forall\epsilon>0,\,\exists\delta>0;\,\forall x\in \Bbb R,\,|x-x_0|<\delta\iff |f(\delta)|\le\epsilon.$$ is not equivalent to continuity.

For example the constant function $f(x)=5$ is continuous and $f(\delta)=5$ for every $\delta $

Thus if your $ \epsilon$ is less than $5$ you can not make $$|f(\delta)|\le\epsilon.$$.

  • $\begingroup$ Right! How about non-constant functions? $\endgroup$ – Kam Apr 8 '18 at 18:29
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    $\begingroup$ @Kam: Still wrong; Say $f(x) = x^2+10$. Then $f(\delta)\geq 10$ for all $\delta$, so you would never have $|f(\delta)|\leq \epsilon$ for most epsilons, but $f$ is certainly continuous everywhere. Your condition “$|f(\delta)|\leq \epsilon$” does not even mention $x_0$; you are trying to claim that continuity at a(n arbitrary point) $x_0$ is equivalent to $f$ taking arbitrarily small values; that’s wrong. And certainly, the fact that $f$ takes small values at some $\delta$ does not tell you two points are $\delta$ close to each other. $\endgroup$ – Arturo Magidin Apr 8 '18 at 18:39

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