Welcome to our community

Be a part of something great, join today!

[SOLVED] Sequential compactness proof

Fermat

Active member
Nov 3, 2013
188
Hello, I'm trying to follow Lemma 12 in the link http://www.econ.brown.edu/fac/Mark_Dean/Maths_RA5_10.pdf.

I can follow it right up until the end, where it asserts that for some N, the ball centred at x_n is contained in a member of O. I do not understand how they got this. Also, for the condradiction wouldn't the radius have to be specifically 1/n, rather than epsilon/2 ?

Thanks
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Hello, I'm trying to follow Lemma 12 in the link http://www.econ.brown.edu/fac/Mark_Dean/Maths_RA5_10.pdf.

I can follow it right up until the end, where it asserts that for some N, the ball centred at x_n is contained in a member of O. I do not understand how they got this. Also, for the condradiction wouldn't the radius have to be specifically 1/n, rather than epsilon/2 ?
So you have this point $x\in S$ (limit of the sequence $(x_n)$), and for some $\varepsilon>0$ the ball $B(x,\varepsilon)$ is contained in the open set $O^*.$ Since $x_n\to x$, there exists $n$ such that $x_n\in B\bigl(x,\frac{\varepsilon}2\bigr)$ whenever $n>N$. It follows (from the triangle inequality) that $B\bigl(x_n,\frac{\varepsilon}2\bigr)\subseteq B(x,\varepsilon)\subseteq O^*$ for all $n>N.$ Finally, you can choose $n$ large enough that $\frac1n<\frac{\varepsilon}2$, and it will then follow that $B\bigl(x_n,\frac1n\bigr) \subseteq B\bigl(x_n,\frac{\varepsilon}2\bigr) \subseteq O^*.$ But that contradicts the fact that $B\bigl(x_n,\frac1n\bigr)$ is not contained in any member of the open cover $O.$

In fact, there is an extra complication here, because $x_n$ belongs to a subsequence of the original sequence $(x^m)$, so you need to think a little in order to convince yourself that this does not seriously affect the proof.
 

Fermat

Active member
Nov 3, 2013
188
Thanks. I assumed \(\displaystyle x^m\) was a typo. Is that incorrect?
Also, any member of the subsequence is a member of the original sequence so that shouldn't be a problem.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Thanks. I assumed \(\displaystyle x^m\) was a typo. Is that incorrect?
Also, any member of the subsequence is a member of the original sequence so that shouldn't be a problem.
The proof starts off with a sequence $(x^m)$, with $B\bigl(x^m,\frac1m\bigr)$ not contained in any set of the covering (the superscript $m$ does not denote a power, of course, it just labels the element of the sequence). The property of sequential compactness then says that this sequence has a convergent subsequence. The usual notation for a subsequence would be to call the $n$th element of the subsequence $x^{m_n}$. But that notation is cumbersome, so the book abbreviates it to $x_n$.

The definition of a subsequence says that the indices $m_n$ of the terms (of the original sequence) that appear in the subsequence should be strictly increasing. This implies that $m_n \geqslant n.$ (If that seems unclear, think about the even numbers as a subsequence of the natural numbers: the $(2n)$th number in the original sequence becomes the $n$th member of the subsequence.) It follows that $B\bigl(x_n,\frac1n\bigr) = B\bigl(x^{m_n},\frac1n\bigr) \supseteq B\bigl(x^{m_n},\frac1{m_n}\bigr),$ which is what is needed for the proof of this result.
 
Last edited: