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The motivation for this question is a comment on this post, consider the logical formula:

$$ ∀\epsilon \in \mathbb{R_{\geq 0}}, ∀x \in \mathbb{R}, ∃δ(x) \in \mathbb{R_{\geq 0}}: |x−x_0|<δ(x)⟹|f(x)−f(x_o)|<ϵ$$

With $f: D \to \mathbb{R}$ where $ D \subset \mathbb{R}$.

Now, apparently any function well defined at $x_o$ is supposed to satisfy this. Well, I can see it should satisfy it when $x=x_o$ but $\forall x$ quantifies for all other $x$s in our domain as well. How would I go about proving that?

Arctic Char
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Clemens Bartholdy
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    As it stands the statement makes no sense, over which set do $x,\epsilon$ range, what is $\delta$ an element of? What is $x_0, l , f$? – jd27 Aug 04 '24 at 08:38
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    Let $f:D\subset \mathbb R\to \mathbb R$. The function $f$ is well defined at a neighborhood of $x_0$ if $\exists \delta >0$ s.t. $]x_0-\delta ,x_0+\delta [\subset (D\cup{x_0})$. For exemple, $f(x)=\frac{1}{x}$ is well defined around $x_0=0$, or $g(x)=\sqrt x$ is not well defined around $x_0=-1$. – Surb Aug 04 '24 at 08:57
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    What is the problem with the questikn – Clemens Bartholdy Aug 05 '24 at 08:21

1 Answers1

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What Did's comment means is that $\forall \varepsilon > 0\; \forall x \;\exists \delta(x) > 0 : 0<|x-x_0|<\delta(x) \implies |f(x) -f(x_0)| < \varepsilon$ is always true. Note the inequalities - if you omit the delta inequality it is trivially always true as you can make the antecedent of the implication false by choosing $\delta(x)=0$, and if you omit the $0<|x-x_0|$ you do not have (a quantifier-swapped variant of) the correct definition of a limit.

The reason it is true for any function is that given $\varepsilon>0$ and $x$, either $|f(x)-f(x_0)|$ is $<\varepsilon$ or it isn't. In the first case we can choose $\delta(x)$ to be anything: any implication with true consequent is true. In the second case, we choose $\delta(x)$ small enough so that $0<|x-x_0| < \delta(x)$ is false. Then the antecedent of the implication is false, so the implication is true.

The thing about well-definedness is a red herring. Every function is well-defined - there's no such thing as a function which is not well-defined.

Matthew Towers
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  • I'd rather say (but of course (s)he should better tell us): "what @Did's comment means is that" $\forall \varepsilon > 0; \forall x ;\exists \delta(x) > 0 : |x-x_0|<\delta(x) \implies |f(x) -f(x_0)| < \varepsilon$ "is always true". No reason to exclude $x=x_0$ in the antecedent: this is the ("quantifier-swapped") definition of continuity at $x_0$. 2) More shortly, "the reason it is true for any function is that given $\varepsilon>0$ and $x$", we can always choose for instance $\delta(x_0)=1$, and $\delta(x)=|x-x_0|/2$ for every $x\ne x_0$.
    • – Anne Bauval Aug 04 '24 at 10:07
  • The comment is talking about limits, not continuity, so I do think the $0<|x-x_0|$ should be there - it just so happens that they're discussing the limit of a continuous function hence the $f(x_0)$ – Matthew Towers Aug 04 '24 at 10:12
  • I agree, but the OP's question is not about the whole post but about point 3 of the comment, where ℓ was replaced by f(x0) . With this replacement, point 3 of the comment was the exact (quantifier swapped) definition of continuity. As for the "correct definition of a limit", it is not unique: https://math.stackexchange.com/questions/1979605 – Anne Bauval Aug 04 '24 at 16:07