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I’m taking my first course in real analysis, and I’m trying to prove the following proposition.

Proposition: If $f:S\to\mathbb{R}$ is uniformly continuous, then $f$ is continuous.

In comparing seemingly similar definitions of continuity and uniform continuity, I noticed that the only difference in the definition is the position of the universal quantification $\forall c\in S$.

  1. $f$ is continuous if $\forall c\in S, \forall \epsilon>0, \exists \delta>0$ such that $\forall x\in S$, if $|x-c|<\delta$ then $|f(x)-f(c)|<\epsilon$.
  2. $f$ is uniformly continuous if $\forall \epsilon>0, \exists \delta>0 \text{ such that } \forall x,c\in S$, if $|x-c|<\delta$ then $|f(x)-f(c)|<\epsilon$.

Leaving only logically relevant symbols,

  1. Continuity: $\forall c \forall \epsilon \exists \delta \forall x P(x, c, \epsilon, \delta)$
  2. Uniform continuity: $\forall \epsilon\exists\delta\forall x \forall c P(x, c, \epsilon,\delta)$

where $P(x,c,\epsilon,\delta)$ is “if $|x-c|<\delta$, then $|f(x)-f(c)|<\epsilon$”

Now, my question is whether it is valid to say that in general:

if $\exists x \forall y P(x,y)$ is true, then so is $\forall y \exists xP(x,y)$.

If this is the case, then I think I can claim that uniform continuity implies continuity, because what I did is essentially to move the $\forall c$ quantification from the right of $\exists\delta$ to its left.

Jaebeom Yim
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    Yes, $\exists x \forall y$ implies $\forall x \exists y$ but not viceversa. If I have a (natural) number $N$ that is "lower-or-equal" of every number, for sure for every number there is a number that is lower-or-equal than it (it is enough to choose the "lowest" one. But the other implications doers not hold: for every (natural) number there is a greater one, but there is no number that is greater than every number. – Mauro ALLEGRANZA Dec 04 '23 at 09:28
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    The proof of the implication that holds is straightforward using logical rules for quantifiers. – Mauro ALLEGRANZA Dec 04 '23 at 09:29
  • I remember when this was taught back when I took analysis. I think it could be enlightening to consider $P(x, y)$ to be a statement about $x, y\in [0,1]$, which means it can be drawn as a subset of the unit square. Then consider what the two statements $\forall x\exists yP(x, y)$ and $\exists y\forall x P(x, y)$ say about said plot. – Arthur Dec 04 '23 at 10:07
  • If $\exists x \forall y P(x,y)$ is true, then so is $\forall y \exists xP(x,y)$. This is true: in the first case same $x$ has to “work” for all $y$, in the second case you get to chose $x$ depending on $y$. This is the reason there are continuous functions that are not uniformly continuous but not the other way around. – Steen82 Jun 14 '24 at 14:04

1 Answers1

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Proof that $\exists x \forall y P(x,y)$ logically implies $\forall y \exists xP(x,y)$

 Let $a$ be arbitrary, and suppose that $\exists x \forall y P(x,y)$ so that we can let $b$ be such that $\forall y P(b,y).$

 As such, we have that $P(b,a).$

 Thus, $\exists xP(x,a).$

 But since $a$ is arbitrary, we have that $\forall y \exists xP(x,y).\;$ QED

P.S. The converse does not hold, since $$\forall y \exists xP(x,y)\iff\forall y P(x_y,y)\kern.6em\not\kern-.6em\implies \exists x \forall y P(x,y).$$

ryang
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