Real Analysis Ep 14: Closed sets 

Please make more youtube courses for other math classes. Ive had an easy time understanding your lectures and they are well done.

I think I've read somewhere that a boundary point x of a set A is a point such that any neighborhood of x contains both points in A, excluding x itself, and points not in A. So this definition would exclude interior points from being limit/accumulation/cluster points. Is this distinction standard in mathematics?

@24:57, I think I was thinking the same (wrong) thing as the student about the limit point question of 0. My idea was pick epsilon to be the closest irrational number to 0. Then since there is an irrational between each two rationals, there couldn't be a rational closer to 0 as then there would be an irrational closer which would contradict that we had picked the closest for epsilon. I suppose this argument fails because we assumed existence of something that does not, in fact, exist. D'oh!

I have a question on closed set F. At 49:12, when we say for any sequence a_n in F, with a_n -> a, then a in F. Here to construct any sequence a_n do we have to find a function f such that f(.): N -> R, where f determines each element of the sequence based on the index "n"? Or we simply find any ordered sequence (by simply checking a_n

Thank you very much,this are very good lectures.

Hi professor, in my multivariable course we briefly mentioned open and closed sets, but we defined them differently. For closed sets, we defined a set F \in R^n to be closed if it contained all of its boundary points, and we defined boundary points using open balls (which I believe are analogous, maybe even entirely equivalent, to epsilon-neighborhoods.) Are there any differences between that definition and the one given in this lecture? I think it's the case that all boundary points are limit points, but not all limit points are boundary points, so it seems the definition I was given is possibly less restrictive, but they seem pretty equivalent to me. (As I think when you remove a limit point from a set, it becomes a boundary point.) Is it more helpful to think of closed sets in one way or another depending on the situation? Again, thanks for the lectures!

Thank u sir. I'm following the lessons well

I think (I may be wrong) that epsilon =x-1 will include the boundary point 1, and so we overshoot the allowed open set condition by the most minimum amount. Doesn't bother me though coz I am an astrophysicist for whom 100 is the same as 999 according to order of magnitude convention!!!!!

The set of all limit points of the set y=2/x+1 :X(-1,1) in R is [1,infinity) but in the set {1/n} why the limit point zero is not included in the set ? Is it because for 1/n is not defined for the limit point zero and y=2/1+x is defined for limit point 1

Sir ,ls there any mathematical reason for the set 1,2,3 is closed

I love you

Q is neiclonorpen