Given a sequence $x_{n}$ in the metric space $(X,d)$, prove that $L$ is a limit point iff there is a subsequence $x_{f(n)}$ which converges to $L$.

Question

Let $(x_{n})_{n=0}^{\infty}$ be a sequence of points in a metric space $(X,d)$, and let $L\in X$. Then the following are equivalent

(a) $L$ is a limit point of $(x_{n})_{n=1}^{\infty}$

(b) There exists a subsequence $(x_{f(n)})_{n=1}^{\infty}$ of the original sequence $(x_{n})_{n=1}^{\infty}$ which converges to $L$.

My solution (Edit)

Let us prove the implication $(a)\Rightarrow(b)$ first.

Indeed, we have the following definition of limit point at hand:

for every $\varepsilon > 0$ and every $N\geq 1$ there is a natural $n\geq N$ such that $d(x_{n},L) < \varepsilon$.

Thus, if we choose $\varepsilon = 1$, there is a natural number $n_{1}\geq 1$ such that $d(x_{n_{1}},L) < 1$.

If we choose $\varepsilon = 1/2$, there is a natural number $n_{2} > n_{1}$ such that $d(x_{n_{2}},L) < 1/2$.

If we choose $\varepsilon = 1/3$, there is a natural number $n_{3} > n_{2}$ such that $d(x_{n_{3}},L) < 1/3$.

Indeed, for every $\varepsilon = 1/j$, there is a natural $n_{j} > n_{j-1} > \ldots > n_{1}$ such that $d(x_{n_{j}},L) < 1/j$.

Taking the limit, we conclude that $x_{f(j)}\to L$, where $f(j) = n_{j}$.

Now it remains to prove the converse implication $(b)\Rightarrow(a)$.

Answer 1

Your proof for "(a) $\implies$ (b)" is incorrect. You have found a bunch of points in the sequence that lie within the distance of $\varepsilon$ of $L$, but that does NOT mean that this subsequence that you constructed converges to $L$. By your construction, all of these points $x_{n_j}$ only satisfy $d(x_{n_j},L)<\varepsilon$ for the same $\varepsilon$. For all we know, it's quite possible that all of them lie more than $\varepsilon/2$ away from $L$. In other words, while all these points lie not to far from $L$, there's no guarantee at all that they are getting closer and closer to $L$.

To fix this flaw, you need to remember that the definition of limits points in metric spaces starts with "For any $\varepsilon>0$, …". And you really need to apply it different values of $\varepsilon$, which you can choose yourself — but make sure to make them smaller and smaller, to enforce the points that you choose to actually keep getting closer and closer to $L$. You can start with "Let $\varepsilon_1=1$; then we can find $x_{n_1}$ such that …". Try to take it from here.

Your reasoning for the other direction is also wrong. You didn't prove anything there … If you construct a correct prove for the first part, hopefully it will help you to start over on this part next. One of your mistakes is exactly the same — you only take one value of $\varepsilon$ (and even worse, you don't explain how you chose it). And then it looks like you picked only two points from the sequence. That doesn't prove anything.

UPDATE: Here's a more detailed explanation of how you can start a proof of "(a) $\implies$ (b)".

• Let $\varepsilon_1=1$ and $N_1=1$. According to the definition of a limit point, there exists an index $n_1\geqslant N_1=1$ such that $d(x_{n_1},L)<\varepsilon_1=1$.

• Let $\varepsilon_2=1/2$ and $N_2=n_1+1$. According to the definition of a limit point, there exists an index $n_2\geqslant N_2$ such that $d(x_{n_2},L)<\varepsilon_2=1/2$.

Can you continue from here?

Answer 2

You already did it here:

Prove that $L$ is a limit point of the sequence $(a_{n})_{n=0}^{\infty}$ iff there is a subsequence $(a_{f(n)})_{n=0}^{\infty}$ which converges to $L$.

But the mistakes in both proofs are the same: for convergence you need to let $\varepsilon\to0$.

Let us correct your first part (as an example), that is, for $(b)\Rightarrow(a)$, following your own text:

Take $n_0\in\mathbb N$ such that $d(x_{n_0},L) < 1$.

Take $n_{1} > n_{0}$ such that $d(x_{n_{1}},L) < \frac12$.

Take $n_{2} > n_{1}$ such that $d(x_{n_{2}},L) < \frac13$.

Proceeding this way, we conclude there exists a subsequence $x_{f(n)}$ which converges to $L$, where $f(j) = n_{j}$.