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So if the definition of continuity is: $\forall$ $\epsilon \gt 0$ $\exists$ $\delta \gt 0:|x-t|\lt \delta \implies |f(x)-f(t)|\lt \epsilon$. However, I get confused when I think of it this way because it's first talking about the $\epsilon$ and then it talks of the $\delta$ condition. Would it be equivalent to say: $\forall$ $\delta \gt 0$ $\exists$ $\epsilon \gt0$ $:|x-t|\lt \delta \implies|f(x)-f(t)|\lt \epsilon$. I guess what I'm asking is whether there is a certain order proofs or more formal statements need to follow. I know I only changed the place where I said there is a $\delta$ but is that permissable in a "formal" way of writing?

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marked as duplicate by user127.0.0.1, mau, Claude Leibovici, vonbrand, Git Gud Mar 10 '14 at 9:41

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up vote 2 down vote accepted

It's already been shown how your definition fails but I'll try to explain why it is the way it is. What the definition tries to get at is basically "You can get as close as you want to the limit." $\epsilon$ represents the any closeness to the limit you want to achieve. and the $\delta$ tells you how to achieve it. In other words, if you're less than $\delta$ distance away from $t$, you'll be less than $\epsilon$ distance from the limit.

Clearly, you need to know how close you want to get at first, that is why $\epsilon$ is chosen first. Hope that helps.

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Perfect I had actually just deduced that exact reasoning myself and was going to ask if that was the reason that's how it is. So as I'm sure you noticed that, that was the definition of uniform continuity, so then for non uniform continuity it's a matter of rearranging where I put each word/symbol/statement? Because we specified that $\epsilon$ was a under a certain condition that implies the uniformness? Does that make sense? – TheHopefulActuary Feb 18 '13 at 3:19
You'd just have to add "$f$ is continuous at $t$ if..." before your given definition. Then everything mentioned in the definition including the choice of $\delta$ is specific to the particular $t$. – genepeer Feb 18 '13 at 3:54
E.g. $f$ is continuous at $1$ if $\forall \epsilon \exists \delta >0 : |x-1|<\delta \implies |f(x)-f(1)|<\epsilon$. – genepeer Feb 18 '13 at 4:01
perfect, thanks! – TheHopefulActuary Feb 18 '13 at 4:03

Let $$f(x) := \begin{cases} 1, & x\in \mathbb{Q} \\ 0, & x\in \mathbb{R} - \mathbb{Q}, \\ \end{cases}$$ a very discontinuous function. Then $\epsilon := 2$ will do, for any $x, t, \delta >0$ you choose. So this cannot be equivalent.

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I assume you meant $\Bbb R\smallsetminus\Bbb Q$, or something like that, rather than $\Bbb R$. – Cameron Buie Feb 18 '13 at 3:09
@CameronBuie: Thanks for catching the typo! – gnometorule Feb 18 '13 at 3:10

What you gave first is the definition of uniform continuity. You have to fix $x$ before embarking the $\forall \epsilon$ thing. That's for continuity at $x$, of course.

Now to answer your question: no, this is not legal to swap $\epsilon$ and $\delta$ like you did.

The funny condition you obtain with this swapping is satisfied by lots of non continuous functions. For instance, any bounded function satisfies it.

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Let's translate it into words, to see how they compare. Continuity at a given point (in this context) means that if we want to keep the $y$-coordinate within a certain interval (no matter how small), then all we have to do is keep the $x$-coordinate from straying too far. (You actually gave the definition for uniform continuity, as it turns out.)

If we simply swap "$\delta>0$" for "$\epsilon>0$" in the definition of continuity, then it means that no matter how far we let the $x$-coordinate stray, the $y$-coordinate will at least manage to stay within some interval (which could be arbitrarily large).

Do you see how (perhaps surprisingly) substantially different the two are?

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