[deleted]

[deleted]

loading...

StepByStep Linear Algebra Solving

Help with graphing circles on Casio CG50

Author : Alfred17686

Subreddit : learnmath

Points : 3

Published : 1 hour ago

Comments : 2

Applied Linear Algebra and Elementary Real Analysis during the summer?

Author : Chem_Whale2021

Subreddit : learnmath

Points : 3

Published : 1 hour ago

Comments : 2

Calculus Integral Equation

Homework help

Existence of the improper integral of (x^a - 1)^(-1/2)

Author : eternalhatesme

Subreddit : learnmath

Points : 3

Published : 2 hours ago

Comments : 5

[College Algebra] Help understanding a function based word problem

Author : ferus_ollin

Subreddit : learnmath

Points : 1

Published : 2 hours ago

Comments : 2

Secret Santa-style probabilities

Author : InternalDecadence

Subreddit : learnmath

Points : 4

Published : 2 hours ago

Comments : 1

How do mathematician generalize concepts?

Author : Marvellover13

Subreddit : learnmath

Points : 8

Published : 2 hours ago

Comments : 5

Why is pi = 3.14159... but in trigonometry itâ€™s equal to 180 degrees??

First existence. Consider balls of radius 1/n centered at p. From that, you get a countably infinite sequence of isolated points X with p as a limit point.

Now to show uniqueness. let q be any point in R^{n} so that q is not p. Let r = |p-q|/2. Then by construction there are infinitely many of X inside B_r(p) and finitely many outside, so there are only finitely many elements of X in B_r(q). Thus you can find R<r so that B_R(q)-q intersects none of X, which means that q is not a limit point.

Let x_n be a sequence in S\{p} which satisfies lim(n to infinity) x_n=p. Since p is a limit point of S, such a sequence must exist. If there is a monotone (increasing or decreasing) infinite subsequence of x_n, then the result will follow almost immediately. So all we need to do is show that such a monotone sequence exists.

Let y_m be the subsequence (possibly of 0 length) of x_n whose elements satisfy x_n>p and let z_k be the subsequence (possibly of 0 length) of x_n whose elements satisfy x_n<p . Then at least one of y_m and z_k must be infinite. Suppose WLOG that it is y_m . Then y_m has limit p. By definition then, for any c>0, there exists M such that m>M implies y_m-p<c. Set b_l by b_l=0.5^(-l). For l=1, find the above M (and call it M1) such that for c=b_1, the above definition holds. Set a_1=y_(M1+1). Set c_l= a_(l-1)-p for l>1 and a_l for l>1 by finding, for c_(l-1), Ml such that the limit definition holds, Ml>M(l-1) and a_l=y_(Ml+1). This recursively defines infinite sequences a_l and c_l. Since a_l is an infinite subsequence of y_m, it surely converges to p. Furthermore, by definition we have a_l -p<c_l =a_(l-1)-p, so a_l <a_(l-1). Then a_l is a monotone decreasing sequence with limit p. Set A to be the collection of elements in a_l. This yields the countable infinite subset whose only limit point is p.

## picado

Hint: construct a sequence with limit p.