Difference between revisions of "1617-257/TUT-R-2"

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(Created page with "We discussed the following on 9/22/16: (1) What are the dimensions of <math>\mathbb{R}^\infty</math> and <math>\mathbb{R}^\omega</math>? (2) Let <math>S</math> be a subset o...")
 
 
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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>).
 
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>).
  
Here's what we would have done if we had extra time to discuss:
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[A student pointed out that I used a definition for limit point which was different from (but also equivalent to) that given in the text. We've replaced any usage of the definition I originally used with the text's definition. We also discussed why the two definitions are equivalent in the Thursday tutorial that took place on 9/29/16.]
  
Let <math>\epsilon > 0</math> be given.
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Let <math>\epsilon > 0</math> be given and let <math>x'' \in S''</math> be given.
  
Let <math>\{x_k\}_{k \in \mathbb{N}} \subset S'</math> be a sequence which converges to <math>x</math>.
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Then there is some <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>.
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There is also a point <math>x \in S</math> such that <math>\|x' - x \| < \epsilon/2.</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>.
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So <math>\|x - x''\| \leq  \|x - x'\| + \|x' - x''\|  < \epsilon.</math>
  
So <math>\|s - x\| \leq \|s - x_n \| + \|x_n - x\| < \epsilon.</math>
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That is, <math>x \in S'.</math>
 
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That is, <math>x \in S'</math>.
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Latest revision as of 12:04, 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).

[A student pointed out that I used a definition for limit point which was different from (but also equivalent to) that given in the text. We've replaced any usage of the definition I originally used with the text's definition. We also discussed why the two definitions are equivalent in the Thursday tutorial that took place on 9/29/16.]

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'.