Quantcast
Channel: Active questions tagged real-analysis - Mathematics Stack Exchange
Viewing all articles
Browse latest Browse all 9155

A simple roadmap to explain the proof in baby rudin theorem 2.43 [closed]

$
0
0

Many have been confused about the proof for theorem 2.43 in baby rudin, and I didn't find it satisfying when I searched it in MSE.Here may serve as a simple explanation:

Two facts:

  1. The initial thought is that for each $x_n$, we shall construct a set $V_n$ which does not contain $x_n$ (you can finetune them lately to get more properties).

  2. A core fact is that for any two points $x_1$ and $x_2$ in $P$, and in any neiborhood of $x_1$, there exists a point $p$ which is diffrent from $x_2$ (no matter $x_1$ is equal to $x_2$ or not), since $P$ is a perfect set (just discuss the radius of this neighbor and compare it to $d(x_1,x_2)$).

The construction process:

  1. For $x_1$, choose any neighbor of it, in this neighbor, find a $p_1$ which is different from $x_1$ and $x_2$, make sure a neighbor of $p_1$, named as $V_1$, doesn't contain $x_1$, this is easy to achieve.

  2. In $V_1$, find a $p_2$ which is different from $x_2$, so we can find a $V_2$, as a neighbor of $p_2$, which does not contain $x_2$. Make sure this $V_2$ is small enough so that $V_2\subset V_1$.

  3. Still, in $V_2$ find a $p_3$ which is different from $x_3$, and find a neighbor of $p_3$ which does not contain $x_3$ and small enough thus $V_3\subset V_2$...

  4. By this process, we can be clear that for any $x_n$, after finite steps a $V_n$ which does not contain it can be constructed, and $V_n\subset V_{n-1}$.

  5. Now let's substitute every $V_n$ by $\overline{V_n}$ , we just make sure every $V_n$ is small enough so this substitution won't destroy the properties we got above.

  6. The following steps are as what Rudin gave.


Viewing all articles
Browse latest Browse all 9155

Trending Articles



<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>