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I am working on a problem in analysis. We are given a sequence $x_n$ of real numbers. Then a definition: A point $c \in \mathbb{R}\cup{\{\infty, -\infty}\}$ is a cluster point of $x_n$ if there is a convergent subsequence of $x_n$ with limit $c$. Then, we let $C$ bet the set of all cluster points. The problem is to prove that the set $C$ is closed. In working on this problem, I tried to determine the cardinality of $C$. And, in doing this I came up with the sequence $x_n = 1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,...$ as an example of a sequence with countable number of cluster points. I believe that for this sequence $C = \mathbb{N}$.

Now consider an enumeration of $\mathbb{Q}$, $q_1, q_2, q_3,...$ And define the sequence $y_n = q_1, q_1, q_2, q_1, q_2, q_3, q_1, q_2, q_3, q_4,...$ It seems to me that the set of cluster points of this sequence is $\mathbb{Q}$.

But now, we could take a sequence from the cluster points of $y_n$ (which is $\mathbb{Q}$) that converges to a member of $\mathbb{Q}^c$ (like $\sqrt{2}$), which would contradict the assumption that the set of cluster points is closed. (Since we have a theorem: a subset of a metric space is closed iff every convergent sequence in the set converges to an element of that set).

So, in trying to prove $C$ is closed, I have come up with what I believe is a counterexample. So my question is: where is the error?

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The error is in the sentence that starts "It seems to me...." The set of cluster points for the sequence you describe is, in fact, all of $\mathbb{R}$, not just $\mathbb{Q}$, for pretty much precisely the reason you give in the next paragraph: for any real number, consider a sequence of rationals converging to that number, and then simply pick a subsequence from your sequence consisting of those rationals.

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  • $\begingroup$ Better is, "consisting of an infinite subsequence of those rationals", I think. $\endgroup$ – Henno Brandsma Feb 5 '16 at 16:44
  • $\begingroup$ Wow I guess that is right. It did not seem possible to me that we could have an uncountable set of cluster points! $\endgroup$ – RJTK Feb 5 '16 at 16:57
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The error is your saying "It seems to me that the set of cluster points of this sequence is $\mathbb Q$" as that is incorrect.

Those are the points which are actually visited an infinite number of times by the points in the sequence. However, there are other points (as you point out with your example $\sqrt{2}$, and in fact all of $\mathbb Q^c$) which are not visited but are approached aribitrarily closely by a subsequence.

The difference is that in the first example, the sequence forms a discrete set (everthing is isolated, all points are bounded apart). That's not the case with the second example.

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  • $\begingroup$ Very nicely answered! Great minds, I see, think alike.... $\endgroup$ – Barry Cipra Feb 5 '16 at 16:39
  • $\begingroup$ Ditto. Will give you +1, @BarryCipra . $\endgroup$ – MPW Feb 5 '16 at 16:39
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For the record, I would like to suggest a topological answer to prove $C$ is closed. Note that $\mathbb{N}$ is closed hence a limit point $x$, of a sequence $f:\mathbb{N}\rightarrow\mathbb{N}$ is in $\mathbb{N}$. The set of cluster points $C$ is equal to the limit points of $f$ restricted to all infinite subsets of $\mathbb{N}$, hence $C\subset\mathbb{N}$ and any subset of $\mathbb{N}$ is closed in $\mathbb{R}$.

Note the same proof does not hold for $f:\mathbb{N}\rightarrow \mathbb{Q}$, since $\mathbb{Q}$ is not closed.

Note the same proof does not hold for $f:\mathbb{N}\rightarrow \mathbb{R}$, since $\mathbb{R}$ does not have a discrete topology.

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