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Let $g(x)$ be: \begin{cases} 0 \space\space \text{if}\space\space x \in \mathbb Q \\ x\space\space \text{if} \space\space x \in \mathbb R - \mathbb Q \\ \end{cases}

I know this function is continuous at $x = 0$ only, so I want to prove, using the negation of $\varepsilon - \delta$ definition, that $g$ is not continuous anywhere else.

$\lim\limits_{x\to a} g(x) \ne g(a)$ means that

$\exists \varepsilon \gt 0, \space\space \forall \delta \gt 0 \space\space$ such that $\space\space$ $0 \lt |x - a| \lt \delta \land |g(x) - g(a)| \ge \varepsilon$

Set $\varepsilon = \frac {|a|} 2$

Let $a \in \mathbb Q$ and select a $x \in (a-\delta, a + \delta) \cap \mathbb R - \mathbb Q$. Such $x$ exists due to irrational numbers being dense in $\mathbb R$.

$|g(x) - g(a)| = |x| \ge \frac {|a|}2$. Since, from the definition, we just need to show that there exists at least one $x$ inside $(a - \delta, a + \delta)$, to satisfy the inequality, choose a $x \gt \frac {|a|}2$.

Thus, $|x| \ge \frac {|a|}2$ holds. There's no rational number, except for $0$, where $g$ is continuous at.

For the second case, let $a \in \mathbb R - \mathbb Q$ and $x \in (a-\delta, a + \delta) \cap \mathbb Q$. The same explanation for the first part can justify the existence of such a $x$.

$|g(x) - g(a)| = |0 - a| = |-a| = |a| \ge \frac {|a|}2$. Then, there's no irrational number where $g$ is continuous at.

Is this correct?

LightL96
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1 Answers1

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You are missing a qualifier that might clear things up. A function $f$ is continuous at a point $a$ if $$\forall \epsilon>0,\exists\delta>0,\forall x\in\text{Domain}(f),0<|x-a|<\delta\implies |f(x)-f(a)|<\epsilon $$ The negation of this is $$\exists \epsilon>0,\forall \delta>0,\exists x\in\text{Domain}(f),0<|x-a|<\delta\ \ \text{and}\ \ |f(x)-f(a)|\ge\epsilon $$ For a proof in the negative, correct, we choose $\epsilon = \frac{|a|}{2}>0$, we then let $\delta>0$ be arbitrary. We then demonstrate that in either case $a\in\mathbb{Q}-{0}$ or $a\in\mathbb{R}-\mathbb{Q}$ we can find an $x$ where $0<|x-a|<\delta\ \ \text{and}\ \ |f(x)-f(a)|\ge\epsilon$. Your proof appears to be correct.

Bobby Ocean
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  • Your negation is incorrect. The correct clause (after the quantifiers) should be $$0 < |x-a| < \delta \quad\text{AND}\quad |f(x)-f(a)| \ge \epsilon$$ – Lee Mosher Feb 21 '24 at 23:15
  • I don't know if there is a logical difference here. In the few real analysis books I have available, I have only ever seen it stated in if/then form. – Bobby Ocean Feb 21 '24 at 23:18
  • To see the logical difference, just compare the truth table of $A \implies (\text{NOT} , B)$ to the truth table of $A , \text{AND} , (\text{NOT} B)$. And if you compare each of those with the truth table of $\text{NOT}(A \implies B)$ you'll find that only one of them is correct, and it's the one I wrote. – Lee Mosher Feb 21 '24 at 23:21
  • What? Truth tables don't exist in 2nd-order logic for statements within scope. I agree a 2nd-order logic statement will collapse to a 1st-order when the scopes are isolated, but that is not the case here. Using $\forall$ and $\exists$ qualifiers do allow the logic of truth tables to come into play. – Bobby Ocean Feb 21 '24 at 23:25
  • I'll say one more thing. You are correct that the proper format for the definition of continuity is a quantified "if-then" statement. But the negation of the definition of continuity is not a quantified "if-then" statement: after swapping all the quantifiers, you then have to negate the "if-then" statement itself, using the standard procedure for negating statements of propositional logic – Lee Mosher Feb 22 '24 at 21:20
  • Here's an MSE question about negating an implication https://math.stackexchange.com/questions/2417770/what-is-the-negation-of-the-implication-statement. – Lee Mosher Feb 22 '24 at 21:23
  • Again, the qualifiers over the scope of the statements can make statements logically equivalent in 2nd order logic. Using 1st order reasoning on statements as a whole without consideration of their scope and meaning just leads to more confusion (in this case we are looking at inequalities bounded over sets of numbers). Recall, 1st order logic has truth tables and True/False; 2nd order does not and requires an "interpretation" before "truth" can have an assignment. – Bobby Ocean Feb 22 '24 at 22:47
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    After some more thought on this, I agree that it should be changed to "AND", since being super formal here it may not be the case that the first property is considered a restraint on the qualifiers set. In practice though, we commonly say things like "there is an $n$, $n>4$ such that $n^2>10$" and maybe even write "$\exists n,n>4\implies n^2>10$". But in this case, it appears the implies is not really an if/then, since the $n>4$ is essentially part of the qualifier of the set of numbers to explore. – Bobby Ocean Feb 22 '24 at 23:04