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December 16th, 2009, 01:19 AM   #1
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Theorem Proof

Let , let , and let be defined by for

(a) If f is continuous at a point , the is continuous at c

(b) If f is continuous on A, then is continuous on A

any help is greatly appreciated.
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December 16th, 2009, 08:18 AM   #2
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Re: Theorem Proof

One of the definitions of continuity is that we can pass the limit into the function. That is a function is continuous at a point a if and only if for every sequence, , such that , we have that . A function is continuous on a subset of the real numbers, A, if and only if the above holds for every a in A. Depending on what you can assume in this proof you might need that definition of continuity. But you can also use the fact that the absolute value of a real number is equal to the square root of the square of that number:

Both the square root function and the function which squares its argument are continuous, so we can use that the composition of continuous functions is continuous. More specifically given two function f: A to B and g: B to C, g(f):A to C is continuous on A if f is continuous on A and g is continuous on B. The square root function is only continuous (or even defined) for nonnegative numbers but that's fine, because the function is continuous for all real x and its range is the nonnegative real numbers, so the absolute value function is continuous on the entire range of
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December 16th, 2009, 10:07 PM   #3
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Re: Theorem Proof

is continuous on , that is the key.
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December 17th, 2009, 08:40 AM   #4
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Re: Theorem Proof

I just said that
forcesofodin is offline  
December 17th, 2009, 12:37 PM   #5
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Re: Theorem Proof

Yeah, but it is much easier to just show absolute value is continous rather then say that it is a composition of continous functions.
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