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Let's say I have a sequence $\{p_n\}$, whose range is the set $E$. Let's say $E$ contains a sequence $\{s_n\}$. Is $\{s_n\}$ necessarily a subsequence of $\{p_n\}$?

The definition of subsequences I must abide by is as follows: For any sequence $\{p_n\}$ in a metric space $X$, let $\{n_k\}$ be a sequence of natural numbers, with $n_1 < n_2 < \ldots < n_k$. Then the sequence given by $\{p_{n_k}\}$ is a subsequence of $\{p_n\}$.

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Consider the sequence $a_n = n$ and $b_n=1$.

In our case $E=\mathbb N$, and clearly $\{b_n\}\subseteq E$, but every subsequence of $a_n$ must have more than two distinct elements.

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    Can the downvoter explain?2012-03-15
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No certainly not. The moment you consider the range set, you lose the order on the terms of the sequence. Examples are easy to come up with.

Let's consider the metric space $\mathbb R$ with the euclidean metric. Consider the sequence

$a_n=\dfrac{1}{2^n}$

Now the range set $E$ of $a_n$ is $E=\left\{\dfrac{1}{2^n}\mid n \in \Bbb N \cup\{0\} \right\}$. Now consider the subsequence that looks like this, $\frac 1 2,1,\frac1 {2^3},\frac 1 {2^2},\frac 1 {2^5},\frac{1}{2^4}, \cdots$

That is the sequence $b_n$ is defined as, $b_n=\begin{cases}\dfrac{1}{2^{n+1}}, \mbox{When } n=2k ~~\mbox{for some} k \in \Bbb N\cup \{0\}\\\dfrac{1}{2^{n-1}}\mbox{When } n=2k+1 ~~\mbox{for some} k \in \Bbb N\cup \{0\}\end{cases}$

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    Hmph, t$r$ue. So$r$$r$y fo$r$ that wrong version. One obvious direction is true, though. :/2012-03-15