Doubts in properties of Sequences

Question

I was studying about limit point of a sequence and in the book it was defined like- "a point will be called as the limit point of a sequence if every neighbourhood of that point will contain an infinite number of members of the sequence" So from this defination of limit point of a sequence can it be claimed that " If a sequence possess a limit point we can find out infinitely many convergent subsequences which converges to the limit point?" And my second question-"Is there any necessity that a sequence must be bounded to have a limit point?"

Answer 1

If a sequence possess a limit point we can find out infinitely many convergent subsequences which converges to the limit point?

This is true. Supposing $\{a_{n_k} \}_{k=1}^\infty$ is a subsequence converging to a point, so too is the subsequence of this made by removing the first point $k=1$. And we can get another subsequence also converging to the same point by then removing $k=2$, etc. In general, we can remove any co-infinite collection of points from a convergent subsequence (not necessarily from the beginning) to get another subsequence also converging to the same point (and there are an infinite number of ways to do this). Herein, we are applying this theorem to the convergent subsequence.

For example, $\{ 1/n\}_{n=k}^\infty$ for any $k \in \mathbb{N}$ is a subsequence of $\{1/n \}_{n=1}^\infty$, and all of these sequences have $0$ as a limit point.

Is there any necessity that a sequence must be bounded to have a limit point?

No need for boundedness. As an example, consider:

$$a_n = \begin{cases} 1/n \qquad \text{If n is odd} \\ n \quad \qquad \text{If n is even} \end{cases}$$

This sequence has $0$ as a limit point but is unbounded since the subsequence of even-indexed terms is unbounded. However, the converse of this statement is true: if $\{a_n\}$ is bounded, then it must possess a limit point; this is the thrust of the Bolzano-Weierstrass theorem and a consequence of the sequential compactness of intervals $[a,b] \subset \mathbb{R}$. As an aside, if we add monotonicity on top of boundedness, then not only does a limit point exist; the whole sequence converges (see: monotone convergence theorem).

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$\underline{\textbf{A Warning}}$:

A convergent sequence $\{a_n\}$ is necessarily bounded. To see why, suppose $a_n \to L$. You can prove that a ball of arbitrary radius centered at $L$ contains all but finitely many points of the sequence. Letting $S$ denote the (finite) set of points in the sequence not contained inside this ball, notice we can fit the entire sequence inside a ball centered at $L$ whose radius is larger than $\max \left(\big\{ |L- x |\big\}_{x \in S} \right)$. This implies that $\{a_n\}$ is bounded.

Answer 2

For the first question the answer is YES. (Just take any subsequence converging to the limit point and the keep omitting the first few terms of the sequence. This gives infinitely many subsequences converging to the limit point). For the second the answer is NO: consider $\{1,\frac 1 2, 2,\frac 1 3 ,3,...\}$. (odd terms are the positive integers and the even terns are $\frac 1 n$). $0$ is a limit point of this sequence.

Answer 3

There are already several good answers to your question, but currently they all construct an infinite set of subsequences just by removing some initial terms for a fixed subsequence. In fact your first statement is true in a much stronger sense: we can find infinitely many subsequences with disjoint indices which tend to a given limit point. This is because if a sequence converges to a given limit then any infinite subsequence also converges to that limit; applying this to a subsequence $a_{n_i}$ which tends to some limit point $x$, we get $$a_{n_{2i+1}}\to x,\quad a_{n_{4i+2}}\to x,\quad a_{n_{8i+4}}\to x,\quad \ldots$$ and no term is used in more than one of these.

Answer 4

Yes you may claim this, that if a sequence $<x_n>$ has a limit point $l$ then there are infinite number of sub sequences $<y_{n_{k}}>$ such that that $<y_{n_{k}}> \rightarrow l$. I think you are well known with this property that if $l$ is a limit point of a sequence then it will have a sub sequence such that this sub sequence will converge to that limit point. Further if we remove some finite number of points from a sequence then it does not affect on its convergence. So, by removing some points again and again from $<y_{n_{k}}>$ we may get infinite number of sub sequences of $<x_n>$ which converges to $l$.

No an unbounded sequence may have limit point. For example you may see the sequence $<x_n>$ defined, $x_n=n,~\text{if n is even}$ and $x_n=1,~\text{if n is odd}$. Here it is clear that sequence $<x_n>$ is unbounded but $1$ is its limit point.