My Math Forum Theorem Proof

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 December 16th, 2009, 01:19 AM #1 Member   Joined: Oct 2009 Posts: 85 Thanks: 0 Theorem Proof Let $A \subseteq \mathbb{R}$, let $f: A \rightarrow \mathbb{R}$, and let $|f|$ be defined by $|f|(x) := |f(x)|$ for $x \in A$ (a) If f is continuous at a point $c \in A$, the $|f|$ is continuous at c (b) If f is continuous on A, then $|f|$ is continuous on A any help is greatly appreciated.
 December 16th, 2009, 08:18 AM #2 Senior Member   Joined: Dec 2009 Posts: 150 Thanks: 0 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, $\{ x_n \| n\in\mathbb{N} \}$, such that $\lim_{n\to\infty} x_n= a$, we have that $\lim_{n\to\infty}f(x_n)= f(\lim_{n\to\infty} x_n) = f(a)$. 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: $|x|= \sqrt{x^2}$ 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 $x\geq 0$ but that's fine, because the function $f(x)= x^2$ 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 $f(x)= x^2$
 December 16th, 2009, 10:07 PM #3 Senior Member   Joined: Jul 2008 Posts: 144 Thanks: 0 Re: Theorem Proof $|x|$ is continuous on $\mathbb{R}$, that is the key.
 December 17th, 2009, 08:40 AM #4 Senior Member   Joined: Dec 2009 Posts: 150 Thanks: 0 Re: Theorem Proof I just said that
 December 17th, 2009, 12:37 PM #5 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 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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### show that the square root function is continuous for non negative reals

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