Re: Question about Contiunuity Definition

On Fri, 23 Nov 2007 18:28:49 -0800 (PST), Venkat Reddy
<vreddyp@xxxxxxxxx> wrote:

On Nov 23, 8:39 pm, LauLuna <laureanol...@xxxxxxxx> wrote:
On Nov 23, 1:29 pm, David C. Ullrich <ullr...@xxxxxxxxxxxxxxxx> wrote:





On Thu, 22 Nov 2007 22:58:58 -0800 (PST), Venkat Reddy

<vred...@xxxxxxxxx> wrote:
On Nov 23, 10:15 am, Tonico <Tonic...@xxxxxxxxx> wrote:
On Nov 23, 6:58 am, Jack <rossb...@xxxxxxx> wrote:

Hi,

I'm using the definition of continuity that says:

let f be a mapping from R into R.

f is continuous at c if:

for every å>0 there exists a ä>0 s.t. |f(x)-f(c)| <å whenever |x-c|<ä and x is in R.

Alright, so my question is, are å and ä, specifically ä, in R as well, or do we not know. I figure it would be stupid if they aren't in R, but do I know for a fact that the ä that exists is in R, or do I have to show that it is/can be?

***************************************************************************---**********
Not only it is clear, imho, that epsilon and delta are reals... where
else could they be? I'm talking of real analysis, of course.

Now, in order to show f is continuous in c you have to prove the
existence of such a delta for any epsilon > 0 whatsoever, but the
"search" is done within R. In this respect you don't know for a fact
that such a delta exists, but you rather prove its existence.

Disclaimer: If you just want to get some answer for your assignment so
that you can get good grades, ignore the following. But if you want to
really think about this, read on.

If you're including "disclaimers" you should begin by specifying
that you're very confused about all this.

Nope. those epsilon and delta do not exist in R. When you try to
verify them by thought process, they practically become infinitesimals
which are "excluded" from R.

This is simply nonsense.

To take a very simple example, say f(x) = x. Say we want to
verify that f is continuous at 0. We'd say this:

Suppose epsilon > 0. Let delta = epsilon. Now if |x - 0| < delta
it follows that |f(x) - f(0)| < epsilon. QED.

That's right but I think Venkat Reddie means that you need a delta for
any possible epsilon, no matter how small the latter is.

So, the existence of the corresponding delta must hold for any term of
an infinite sequence of epsilons with limit 0. Isn't this the gist of
the question?

Of course, none of the epsilons is neither 0 nor infinitesimal; this
is what Venkat should consider here.

Ok. You mean epsilon will never reach the status of the "infintely
small" number while checking for continuity (even by logic),

I'm not sure who this is addressed to, nor what anyone may or may
not mean. But in fact _yes_ that's correct, in the definition
of continuity epsilon never "reaches the status of" the infinitely
small.

Mathematical quantities do not "reach" any sort of "status".

Definition mean exactly what they say they mean. The definition
of continuity says "For every (real) epsilon > 0 there exists
a (real) delta > 0 such that...". Nothing there about anything
"reaching" the "status" of "infinitely small".

but just
represents "any" number between 0 and |x-c|.

No, epsilon represents any positive real number!

May be you are right. I'm
not sure.

Well that's progress. All you need to do is learn to
read exactly what mathematical statements say and
you'll be much less confused about a lot of things.

- venkat





************************

David C. Ullrich
.

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