continues prove question..

**transgalactic** · December 23rd 2008, 11:48 AM

prove that all the points of discontinuity
of a monotonic and bounded function
are points of discontinuity of the first type?

**TwistedOne151** · December 23rd 2008, 01:09 PM

Recall that a point x=c is a discontinuity of the first kind if $\lim_{x\to{c-}}f(x)$ and $\lim_{x\to{c+}}f(x)$ exist. Suppose that f(x) is monotonically increasing (an analogous argument will exist for monotonically decreasing f(x)). Then for an increasing infinite sequence of $x_n<c$ with $\lim_{n\to\infty}x_n=c$ , we see that $f(x_n)$ is an increasing sequence. Consider this in combination with the boundedness of f(x), and what that says about the existence of $\lim_{x\to{c-}}f(x)$ . Similar work for a decreasing sequence $x_n>c$ , $\lim_{n\to\infty}x_n=c$ will give a similar result for $\lim_{x\to{c+}}f(x)$ .

--Kevin C.

**transgalactic** · December 26th 2008, 12:03 PM

a monotonic function is a function which is systematically increasing or decreasing.

a bounded function means that there is at least one subsequence
which converges to limit.

so our function converges if its monotonic and bound.
and every subsequence converge to the same limit.

so you say that this limit is a point of discontinuity called C.

and we have two cases
the first is for monotonically increasing f(x)

and that it converges to point C from the left (minus side)

but C cannot be the limit of the function
after a discontinuity point

the function must go on and converge to real limit
it cannot stop on point C

??

**Mathstud28** · December 26th 2008, 03:09 PM

To OP: Possible alternate approach...wait for someone to confirm.

You could also approach this from the contradiction approach. Let $f$ be a montonically increasing real function $f$ . Next suppose that $c$ is a discontinuity of $f$ but it is of the second kind. This implies that $\lim_{x\to{c^+}}f(x)$ or $\lim_{x\to{c^-}}f(x)$ does not exist. I will show case of $\lim_{x\to c^-}f(x)$ and you can extrapolate the other case. So suppose that $\lim_{x\to c^-}f(x)$ does not exist. This implies that there does not exist a $p>0$ ( the case is similar for negative p) such that for every $\varepsilon>0$ there exist a $\delta>0$ such that $c-x<\delta\implies |p-f(x)| <\varepsilon$ . So for any $\delta$ we can find a $x_1<c$ such that $c-x_1<c-x<\delta$ but $|p-f(x_1)|>|p-f(x)|\implies f(x_1)<f(x)$ , but this violates that $f$ is monotonically increasing.

**transgalactic** · December 27th 2008, 05:51 AM

$\varepsilon>0$ mean that on certain surrounding

so $\delta>0$

what is the meaning of $\delta>0$ ??

c is our point of discontinuity
what is x??

why
$c-x<\delta\implies f(c) - f(x) <\varepsilon$

??

**transgalactic** · January 7th 2009, 11:23 AM

by this
$ c-x<\delta\implies |p-f(x)| <\varepsilon $
this is a definition of a limit

can you explain in simple words
why are doing this
what is the meaning of this part

$ c-x_1<\delta $
but
$ |p-f(x_1)|>|p-f(x)|\implies f(x_1) $

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