Can someone with the book ready to hand explain how he proves the converse? He just says taking $\displaystyle {\delta_n}=1/n$, we thus find a sequence......

Sometimes he's a little too presuming of the reader's perception.

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- Oct 17th 2011, 11:41 AMboromirTheorem 4.2 in Rudin POMA
Can someone with the book ready to hand explain how he proves the converse? He just says taking $\displaystyle {\delta_n}=1/n$, we thus find a sequence......

Sometimes he's a little too presuming of the reader's perception. - Oct 17th 2011, 11:55 AMgirdavRe: Theorem 4.2 in Rudin POMA
Thanks to this choice, we can find a sequence $\displaystyle \{x_n\}$ such that $\displaystyle d(x_n,p)\leq\frac 1n$ and $\displaystyle d(f(x_n),f(p))\geq \varepsilon$, hence we have shown that we don't have the part after the "if and only if".

- Oct 17th 2011, 11:58 AMPlatoRe: Theorem 4.2 in Rudin POMA
You may be mixing apples and oranges.

He means for each $\displaystyle \delta_n=\tfrac{1}{n}$ we can find a $\displaystyle p_n$ such that $\displaystyle d(p_n,p)<\delta_n$ but $\displaystyle d(f(p_n),q)\ge\epsilon $.

So $\displaystyle (p_n)\to p$ but $\displaystyle f(p_n)\not\to q$. - Oct 17th 2011, 12:10 PMboromirRe: Theorem 4.2 in Rudin POMA
I think you mean q instead of F(p). So essentialy that is chosen since 1/n-> 0 (allowing x_n to be arbitarily close to 0) and x_n consists of all 'rogue' values of x is so far as they make F(x) more than epsilon away from q. Is this correct?

- Oct 17th 2011, 12:16 PMPlatoRe: Theorem 4.2 in Rudin POMA