Show that lim S is a closed set

[Mr Davis 97]

Homework Statement

Suppose that p_n → p and each p_n lies in lim S. We claim that p ∈ lim S. Since
p_n is a limit of S there is a sequence (p_n,k)_k∈N in S that converges to p_n as k →∞.
Thus there exists q_n = p_n,k(n)∈ S such that
d(p_n, q_n) <1/n.
Then, as n→∞ we have
d(p, q_n) ≤ d(p, p_n) + d(p_n, q_n) → 0
which implies that q_n → p, so p ∈ lim S, which completes the proof that lim S is a
closed set.

Homework Equations

The Attempt at a Solution

The only part of the proof I don't understand is the following:

Since
p_n is a limit of S there is a sequence (p_n,k)_k∈N in S that converges to p_n as k →∞.
Thus there exists q_n = p_n,k(n)∈ S such that
d(p_n, q_n) <1/n.

Where does the 1/n come from? Why can we be sure that such a q_n exists?

 

[Dick]

Homework Statement

Suppose that p_n → p and each p_n lies in lim S. We claim that p ∈ lim S. Since
p_n is a limit of S there is a sequence (p_n,k)_k∈N in S that converges to p_n as k →∞.
Thus there exists q_n = p_n,k(n)∈ S such that
d(p_n, q_n) <1/n.
Then, as n→∞ we have
d(p, q_n) ≤ d(p, p_n) + d(p_n, q_n) → 0
which implies that q_n → p, so p ∈ lim S, which completes the proof that lim S is a
closed set.

Homework Equations

The Attempt at a Solution

The only part of the proof I don't understand is the following:

Since
p_n is a limit of S there is a sequence (p_n,k)_k∈N in S that converges to p_n as k →∞.
Thus there exists q_n = p_n,k(n)∈ S such that
d(p_n, q_n) <1/n.

Where does the 1/n come from? Why can we be sure that such a q_n exists?

You are given that ##p## is a limit of points of ##lim S##. You want to show that ##p## itself is an element of ##lim S##. To do this you have show that there is a sequence of point in ##S## whose limit is ##p##. The argument shows that for every ##n## there is a point of ##S## whose distance from ##p## is less than ##1/n##. Which goes to zero as ##n \rightarrow \infty##. Hence ##p## is in ##lim S##. Clearer?

 

[FactChecker]

Where does the 1/n come from?

Anything that goes to 0 will do the trick. You can use 1/n and n→∞ or you can use ε>0 and ε→0. They will both make the distance go to 0.

Why can we be sure that such a q_n exists?

Because p_n is a limit of points in S, there is a q_n at least that close to p_n.

 

[Mr Davis 97]

I guess my main confusion lies in the following: Thus there exists q_n = p_{n,k(n)∈ S} such that d(p_n, q_n) <1/n.

Does this somehow come from the definition of convergence? Also, I am confused by the notation. The proof says that there is a sequence that converges to p_n. So are we saying that n is some fixed number in this instance? Because obviously a sequence can't converge to another sequence.

 

[FactChecker]

I guess my main confusion lies in the following: Thus there exists q_n = p_{n,k(n)∈ S} such that d(p_n, q_n) <1/n.

Does this somehow come from the definition of convergence?

Yes. It comes from the sentence before: p_n is a limit of S there is a sequence (p_n,k)_k∈N in S that converges to p_n as k →∞.

Also, I am confused by the notation. The proof says that there is a sequence that converges to p_n. So are we saying that n is some fixed number in this instance?

For that statement, n is fixed. But the logic will work for any n, which is important since n →∞.

Because obviously a sequence can't converge to another sequence.

I agree. and it doesn't even have to converge to the same limit as the other sequence. It is converging to the limit of the points, p_n that the sequences (p_n,k)_k∈N converge to.

 

[FactChecker]

Start with the blue sequence of points, p_n, converging to the red point, A. Each blue point is in S, so there is a sequence of green points converging to each one. Select the sequence of circled points converging to A. That circled sequence proves that A is also in S.