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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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up vote 2 down vote accepted

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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+1. Nice and short answer. : ) – user21436 Mar 15 '12 at 0:50
Can the downvoter explain? – Asaf Karagila Mar 15 '12 at 18:53

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


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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Would you mind providing please one or two such examples? – jamaicanworm Mar 15 '12 at 0:23
I have added an example, is that fine? – user21436 Mar 15 '12 at 0:35
Your added statement is false. Think about $a_n=0$ for even $n$ and $a_n=n$ otherwise. It has a constant subsequence but the range is infinite. – Asaf Karagila Mar 15 '12 at 0:45
Hmph, true. Sorry for that wrong version. One obvious direction is true, though. :/ – user21436 Mar 15 '12 at 0:47

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