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Say I'm given the sequence $\{a_1,a_2,a_3,\dots\}$. Does a subsequence have to be infinite? Or can it be finite too? For example, is $\{a_1,a_2,a_3\}$ a subsequence?

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It depends on the context. –  Mariano Suárez-Alvarez Feb 23 at 9:10
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Yes the subsequence must be infinite. Any subsequence is itself a sequence, and a sequence is basically a function from the naturals to the reals.

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Usually, this is the definition of subsequence.

Definition. Let $\{p_n\}_n$ be a sequence (in some set), and let $n_1<n_2<n_3<\ldots$ be a strictly increasing sequence of positive integers. Then the sequence $\{p_{n_j}\}_j$ is called a subsequence of $\{p_n\}_n$.

It follows that $\{p_1,p_2,p_3\}$ is not a subsequence of $\{p_n\}_n$, since $n_1=1$, $n_2=2$, $n_3=3$, but what is $n_4$? This is the common feeling for mathematical analysts.

However, we must be careful, since it all boils down to the very definition of sequence. In analysis, a sequence is a function $\mathbb{N} \to X$, or, more generally, a function $N \to X$, where $N \subset \mathbb{N}$ is such that $N$ contains all sufficiently large integers. In other disciplines, it might be useful to call sequence any function defined on a subset of $\mathbb{N}$, even a finite one.

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A sequence in a set $X$ is a function $f:\mathbb{N}\rightarrow X$. Let $g:\mathbb{N} \rightarrow \mathbb{N}$ be a strictly increasing sequence, then the composition of $f$ and $g$, is called a subsequence of $f$. Thus the domain of a subsequence is always infinite, but the range can certainly be finite.

As an example consider the sequence $f(n):=n$ even $n$ and $f(n):=1$ odd $n$. Let $g(n):=2n+1$. The range of $f\circ g$ is equal to $\{1\}$, which is finite.

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In my experience, the term subsequence has always referred to an infinite subsequence. I believe this is standard. If you want to consider finite subsequences, you should say finite subsequence, or if you want to consider subsequences that may or may not be finite, you can say possibly finite subsequence.

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A subsequence can be finite. Note however, that you can always extend to an infinite sequence by $a_1, a_2, a_3, a_3, a_3, \cdots$

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I really don't think $a_1,a_2,a_3,a_3,\ldots$ is typically considered a subsequence of $\{a_i\}_{i=1}^\infty$. –  froggie Feb 23 at 9:10
    
-1 $a_1, a_2, a_3, a_3, a_3, \cdots$ is not a subsequene –  miracle173 Feb 23 at 9:12
    
I never said that it was a subsequence. I said that if you want to extend a finite sequence (which means it converges!) to an infinite sequence you can do so by simply extending it trivially to the convergent value. Why the downvotes? –  Euler....IS_ALIVE Feb 23 at 9:13
    
This is quite ridiculous. Does the sequence $(1,1,1,1,1 \cdots)$ have a convergent subsequence? Of course it does; namely the sequence $1$. –  Euler....IS_ALIVE Feb 23 at 9:15
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@Euler....IS_ALIVE, I am sorry but that is ridiculuous in the usual meaning of subsequences. If that were a sensible interpretation of the term subsequence, then there would be no point in Bolzano's theorem, for all sequences $a_1$, $a_2$, $\dots$ would have a convergent subsequence, whether they be bounded or not: namely $a_1$. –  Mariano Suárez-Alvarez Feb 23 at 9:57
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