# Thread: Theorem 4.2 in Rudin POMA

1. ## Theorem 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.

2. ## Re: 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".

3. ## Re: Theorem 4.2 in Rudin POMA

Originally Posted by boromir
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......
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 $\displaystyle f(p_n)\not\to q$.

4. ## Re: 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?

5. ## Re: Theorem 4.2 in Rudin POMA

Originally Posted by boromir
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?
The sequence $\displaystyle (p_n)$ is close to $\displaystyle p$ not $\displaystyle 0$.
But the sequence $\displaystyle f(p_n)$ is not close to $\displaystyle q$.