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1  Mathematics / Real Analysis Forum / Re: Please help me with this inequality on: April 29, 2010, 01:05:33 AM
Ok, I solved the problem.
2  Mathematics / Real Analysis Forum / Please help me with this inequality on: April 27, 2010, 11:24:34 PM
http://data.artofproblemsolving.com/aops20/latex/texer/c4def8b9eded12bf9c988beb3aadfed1674fbb2b.png

My question is how equation 1 equals equation 2?
Thanks.
3  General Category / General Discussion / A website for you to edit math formula online. on: March 23, 2010, 09:26:21 PM
http://www.codecogs.com/components/equationeditor/equationeditor.php

I hope mathcs.org will support LaTex
4  Mathematics / Real Analysis Forum / some typos on: August 20, 2009, 04:07:04 AM
http://www.mathcs.org/analysis/reals/integ/techniqs.html

Example 7.2.2: Standard Antiderivatives

(g), (h), (i) have 3 typos, one dx should be 1.
5  Mathematics / Real Analysis Forum / Rolle's Theorem on: August 14, 2009, 03:06:29 AM
http://www.mathcs.org/analysis/reals/cont/proofs/rollethm.html

"Assume for now that f(c) # 0 is a maximum. Since f(a) = f(b) = c we know that c is in (a, b), and therefore f is differentiable at c. Note that f(x) f(c) since f(c) is a maximum. "

I think f(a)=f(b)=c  is a typo, it should be f(a)=f(b)=0.

But why c is in(a,b)  then we can get that f is differentiable at c? please explain.

6  Mathematics / Real Analysis Forum / Definition of preimage or inverse image on: August 10, 2009, 02:54:43 AM
For discussion.

Let A and B be two sets and f a function from A to B. If C is a subset of the range B then the preimage, or inverse image, of C under the function f is the set defined as f -1(C) = {x belong to A : f(x) belong to C }

http://www.mathcs.org/analysis/reals/logic/relation.html

I think C is not necessary a subset of the range B. C can be any set as long as" f -1(C) = {x belong to A : f(x) belong to C } "

Otherwise, we can not solve this question by "http://www.mathcs.org/analysis/reals/cont/answers/topcont2.html"
Let f(x) = 1 if x > 0 and f(x) = -1 if x  0. Show that f is not continuous by
1.finding an open set whose inverse image is not open.
2.finding a closed set whose inverse image is not closed.

the range of f(x) is 1 and -1, and are only these two numbers, but (-2,0) is not a subset of B. If we use above inverse image definition even the question "find an open set" itself is not correct.
So I think the domain of inverse image is not necessarily a subset of rang f.
7  Mathematics / Real Analysis Forum / Re: Proposition 6.4.1: Continuity and Topology on: August 05, 2009, 10:13:25 PM
Thanks.
8  Mathematics / Real Analysis Forum / Proposition 6.4.1: Continuity and Topology on: August 04, 2009, 11:35:48 AM
http://www.mathcs.org/analysis/reals/cont/proofs/conttop.html

For the proof (2) => (1).

(2) => (1): Assume that the inverse image f-1(U) of every open set U is open.

Take any point x0 (belong to ) "U" and pick an ε > 0.

Should  it be : take any point x0 (belong to) "D", insteand belong to "U"?
9  Mathematics / Real Analysis Forum / Re: Another question on: August 04, 2009, 10:42:36 AM
Thanks, I see.
10  Mathematics / Real Analysis Forum / Another question on: August 01, 2009, 11:18:26 AM
http://www.mathcs.org/analysis/reals/cont/proofs/monodisc.html

From last paragraph.

"Since the sum of jumps must be smaller than f(b) - f(a), the set J(n) is finite for all n".

I can uderststand that the sum of jumps must be samller than f(b)-f(a), but why the Set J(n) is finite then?

Thanks.
11  Mathematics / Real Analysis Forum / A question on: August 01, 2009, 11:14:02 AM


http://www.mathcs.org/analysis/reals/cont/answers/discwp1.html

"We can easily check that the limit as x approaches 3 from the right and from the left is equal to 4. Hence, the limit as x approaches 3 exists, and therefore the function has a removable discontinuity at x = 3. If we define k(3) = 4 instead of k(3) = 1 then the function in fact will be continuous on the real line."

I think from the left is eqal to 6 not to 4. Thanks.

I love your real analysis website.
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