limit of subsequence where $X_n - X_{n-1}\rightarrow 0$. suppose $X_{n}$ is a sequence of real numbers such that $X_{n} - X_{n-1} \rightarrow 0$.
prove that the limit of subsequence is empty or single point set or interval.
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I know the limit of subsequence is the set of limits of subsequences of {$P_{n}$} n=1,2,...
{$P_{n}$} is the sequence in metric space (X,d).
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My effort in this regard is as follows:
Suppose it has two boundary points.
We want to prove that all points between these two points are boundary points.such as a & b.
Consider a point between these two points.such as c.(a<c<b)
Now we have to consider the radius of the neighborhood around this point.
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Now I can not calculate this radius correctly and I do not know how the rest of the question will be proved.
Please help me!
 A: Suppose there are $J$ and $M$ such that $J<M$ and they both are limits of some subsequences. Suppose then, there exist $K$ and $L$ such that $J<K<L<M$ and no $x\in (K,L)$ is a limit point of any subsequence. This contradicts assumptions:
if $J$ is a limit point of some subsequence, then there must be infinitely many terms in $(X_i)_{i\in \mathbb N}$ belonging to any, arbitrarily small neighborhood of $J$;
similary there must be infinitely many terms arbitrarily close to $M$;
so, frankly speaking, the sequence must at least bounce between $J$ and $M$, and it must do so infinitely many times;
however, convergence of differences to $0$ implies that for any $\varepsilon>0$ there exists an index $m$ such that for each $i>m$ differences are smaller than $\varepsilon$: $|X_i-X_{i-1}|<\varepsilon$;
but for $\varepsilon < L-K$ it would force some – and actually infinitely many – terms of the sequence to fall inside the forbidden $(K,L)$ interval.
So, if there exist subsequences in $(X_i)_{i\in \mathbb N}$ with different limits $J<M$, then each point of interval $[J,M]$ is an accumulation point of the sequence, in other words terms of $(X_i)$ are dense in the interval, hence each point in the interval is a limit of some subsequence.
A: So suppose that $a$ and $b$ are limits of subsequences of $(x_n)_n$ and that $c \in (a,b)$.
We want to prove that there exists a subsequence $(x_{n_k})_k$ of $(x_n)_n$ such that $x_{n_k} \rightarrow c$.
Choose $\varepsilon = 1/k$. Then there exists an integer $N$ such that $(\forall n > N)(|x_n - x_{n-1} | \leq \varepsilon)$. Since $(x_n)_n$ has subsequences converging to $a$ and $b$ we can choose integers $m > N$ and $p > N$ such that $|x_m - a| < \varepsilon$ and $|x_p - b| < \varepsilon$, and furthermore $c \in (x_m,x_p)$ (because we can choose $x_m$ and $x_p$ arbitrarily close to $a$ and $b$ respectively).
Assume now that $m<p$ (the case $m>p$ is completely analogous). Then the sequence $x_m, x_{m+1}, \dots, x_p$ takes steps smaller then $\varepsilon$ between consecutive numbers, and goes from $x_m < c$ to $x_p > c$. Therefore it contains some number $x_q$ such that $|x_q - c| < \varepsilon$ (in fact even $|x_q - c| < \varepsilon/2$).
Note that we can make $N$ as large as we want. Hence, if we make a sequence $x_{n_1}, ..., x_{n_{k-1}}$ with $|x_{n_i} - c| < 1/i$, we can choose $N > x_{n_{k-1}}$ to choose a $x_{n_k}$ greater than $x_{n_{k-1}}$ and $|x_{n_k} - c| < 1/k$. This way one can construct a subsequence of $(x_n)_n$ converging to $c$.
A: Let $\varepsilon > 0$ and $n_0 \in \mathbb N$. Let $(x_{n_k})_k$ be a subsequence of $(x_n)$ with $x_{n_k} \to a$ and $(x_{m_j})_j$ one that converges to $b$. Now let $j , k \in \mathbb N$ such that $n_k < m_k$, $x_{n_k} < c < x_{m_j}$, and $\lvert x_{n+1} -x_n \rvert < \varepsilon$ for all $n > n_{k}$. Then there is an index $\tilde n > n_k$ with $x_{\tilde n} \le c \le x_{\tilde n + 1}$ or in other words $$c - x_{\tilde n} \le x_{\tilde n +1 } - x_{\tilde n} < \varepsilon .$$
Therefore $c$ is an accumulation point of $(x_n)$.
