Theorem 4.2 in Rudin POMA

**boromir** · October 17th 2011, 11:41 AM

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

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

**girdav** · October 17th 2011, 11:55 AM

Thanks to this choice, we can find a sequence $\{x_n\}$ such that $d(x_n,p)\leq\frac 1n$ and $d(f(x_n),f(p))\geq \varepsilon$ , hence we have shown that we don't have the part after the "if and only if".

**Plato** · October 17th 2011, 11:58 AM

Originally Posted by boromir

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

You may be mixing apples and oranges.
He means for each $\delta_n=\tfrac{1}{n}$ we can find a $p_n$ such that $d(p_n,p)<\delta_n$ but $d(f(p_n),q)\ge\epsilon$ .
So $(p_n)\to p$ but $f(p_n)\not\to q$ .

**boromir** · October 17th 2011, 12:10 PM

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?

**Plato** · October 17th 2011, 12:16 PM

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?

Look at reply #3.
The sequence $(p_n)$ is close to $p$ not $0$ .
But the sequence $f(p_n)$ is not close to $q$ .

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