10-327/Classnotes for Thursday November 4

DBN: Classes: 2010-11: 10-327/Navigation Additions to the MAT 327 web site no longer count towards good deed points # Week of... Notes and Links 1 Sep 13 About This Class, Monday - Continuity and open sets, Thursday - topologies, continuity, bases. 2 Sep 20 Monday - More on bases, Thursdsay - Products, Subspaces, Closed sets, HW1, HW1 Solutions 3 Sep 27 Monday - the Cantor set, closures, Thursday, Class Photo, HW2, HW2 Solutions 4 Oct 4 Monday - the axiom of choice and infinite product spaces, Thursday - the box and the product topologies, metric spaces, HW3, HW3 Solutions 5 Oct 11 Monday is Thanksgiving. Thursday - metric spaces, sequencial closures, various products. Final exam's date announced on Friday. 6 Oct 18 Monday - connectedness in ${\mathbb {R} }$, HW4, HW4 Solutions, Thursday - connectedness, path-connectedness and products 7 Oct 25 Monday - Compactness of $[0,1]$, Term Test on Thursday, TT Solutions 8 Nov 1 Monday - compact is closed and bounded, maximal values, HW5, HW5 Solutions, Wednesday was the last date to drop this course, Thursday - compactness of products and in metric spaces, the FIP 9 Nov 8 Monday-Tuesday is Fall Break, Thursday - Tychonoff and a taste of Stone-Cech, HW6, HW6 Solutions 10 Nov 15 Monday - generalized limits, Thursday - Normal spaces and Urysohn's lemma, HW7, HW7 Solutions 11 Nov 22 Monday - $T_{3.5}$ and $I^{A}$, Thursday - Tietze's theorem 12 Nov 29 Monday - compactness in metric spaces, HW8, HW8 Solutions, Thursday - completeness and compactness 13 Dec 6 Monday - Baire spaces and no-where differentiable functions, Wednesday - Hilbert's 13th problem; also see December 2010 Schedule R Dec 13 See December 2010 Schedule F Dec 20 Final exam, Monday December 20, 2PM-5PM, at BR200 Register of Good Deeds Add your name / see who's in! See Hilbert's 13th

See some blackboard shots at BBS/10_327-101104-142342.jpg.

Dror's notes above / Student's notes below

• Question: Regarding the proof of the Lebesque Number Lemma, let T(x) be the function we were working with in the proof. I am confused with how we reached the conclusion that if d(x,y)<E, then T(y)>= T(x) - E. I know that it was said that this is just an application of the triangle inequality, but I am having a bit of trouble seeing that. Hopefully someone can make this point a bit clearer for me. Thanks! Jason.
  • If you could find a ball of radius 7 around $x$ which fits inside some set $U$, and you move $x$ just a 1 unit away to $y$, then by the triangle inequality the ball of radius 6 around $y$ is entirely contained inside the ball of radius 7 around $x$ so it is entirely contained in $U$. Drorbn 18:37, 6 November 2010 (EDT)
    • I have some doubts with Lebesgue number lemma too.. this delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem? Don't we need to subtract some small positive value and then find a valid radius? I am not sure if there should be some technicality involved here.-Kai

And once we found this delta(x_0) we should divide by 2 so that delta(x)>delta(x_0) for all x in X? Because it is not necessarily true that we can find a ball of radius delta(x_0) around x_0 to fit into ball?-Kai

• I think you are correct, (if you follow the proof on the blackboard shots to the letter) as an example of what you are talking about take the open cover of [0,1] using sets of the form [0,x) for x<3/4 and (y,1] for y>1/4. Then $\Delta (x)\geq 1/4$ and at 1/2 $\Delta (1/2)=1/4$ but no ball of radius 1/4 around 1/2 fits inside any of the sets. So we will need to take a smaller value than $inf(\Delta (x))$ as our value of $\delta _{0}$ for Lebesgue's Lemma. Dividing by two as you suggest should work fine to fix this problem... Maybe it's just me, but in proof's like this I always feel the urge to divide the final answer by 2, just in case I mixed up some some strict inequality, with a non-strict one somewhere - John
  • Thanks John very nice example. Can you also help me with this question? "delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem in showing delta(y)>=delta(x)-epsilon when d(x,y)<epsilon?"-Kai
    • I don't think there is any problem with this step.

Without loss of generality $\epsilon <\Delta (x)$ otherwise the condition holds vacuously. Suppose we construct a ball of radius r such that $r>\epsilon$ around x so that it is a subset of some U in the cover, then we can construct a ball of radius $r-\epsilon$ around y such that this ball is also in U. So $\Delta (y)\geq r-\epsilon$ for all r such that a ball of radius r exists around x as in the proof. Which implies $\Delta (y)\geq sup(r)-\epsilon$ implies $\Delta (y)\geq \Delta (x)-\epsilon$. Hopefully this makes sense and works - John -Yes. Makes sense. Nice and concise thanks! -Kai