1617-257/TUT-R-2: Difference between revisions

Revision as of 10:57, 30 September 2016

We discussed the following on 9/22/16:

(1) What are the dimensions of $\mathbb {R} ^{\infty }$ and $\mathbb {R} ^{\omega }$ ?

(2) Let $S$ be a subset of $\mathbb {R} ^{n}$ . Show that the set of limit points of $S$ , $S'$ , is closed.

A student gave an for problem (2) which works fine if $S$ is a closed set (it depended on the fact that $S'\subset S$ ).

Let $\epsilon >0$ be given and let $x''\in S''$ be given.

Then there is some $x'\in S'$ such that $\|x''-x'\|<\epsilon /2.$

There is also a point $x\in S$ such that $\|x'-x\|<\epsilon /2.$

So $\|x-x''\|\leq \|x-x'\|+\|x'-x''\|<\epsilon .$

That is, $x\in S'.$

@@ Line 11: / Line 11: @@
 A student gave an for problem (2) which works fine if <math>S</math> is a closed set (it depended on the fact that <math>S' \subset S</math>).
+Let <math>\epsilon > 0</math> be given and let <math>x'' \in S''</math> be given.
-Here's what we would have done if we had extra time to discuss:
-Let <math>\epsilon > 0</math> be given.
+Then there is some <math>x' \in S'</math> such that <math>\|x'' - x'\| < \epsilon/2.</math>
-Let <math>\{x_k\}_{k \in \mathbb{N}} \subset S'</math> be a sequence which converges to <math>x</math>.
+There is also a point <math>x \in S</math> such that <math>\|x' - x \| < \epsilon/2.</math>
-Then there is some <math>n</math> for which <math>\| x_n - x\| < \epsilon/2</math>.
+So <math>\|x - x''\| \leq  \|x - x'\| + \|x' - x''\|  < \epsilon.</math>
+That is, <math>x \in S'.</math>
-Since <math>x_n \in S'</math>, there is some <math>s \in S</math> such that <math>\|x_n - s\| < \epsilon/2</math>.
-So <math>\|s - x\| \leq \|s - x_n \| + \|x_n - x\| < \epsilon.</math>
-That is, <math>x \in S'</math>.