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Let $a$ be an real number and let $S$ be the set of all sequences in $\mathbb{R}$ converging to $a$. What is the Cardinality of $S$?

Thanks

1 Answers 1

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First note that there are only $2^{\aleph_0}$ sequences of real numbers. This is true because a sequence is a function from $\mathbb N$ to $\mathbb R$ and we have $$\left|\mathbb{R^N}\right|=\left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$$

Now note that take any injective sequence which converges to $a$, then it has $2^{\aleph_0}$ subsequences. All are convergent and they all converge to $a$.

Therefore we have at least $2^{\aleph_0}$ sequences converging to $a$, but not more than $2^{\aleph_0}$ sequences over all, so we have exactly $2^{\aleph_0}$ sequences.

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    Thanks @AsafKaragila. Just one Question: The set of all convergent sequences in $\mathbb{R}$ have this same cardinality?2012-10-22
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    @Tomás: The same argument applies. We know that there are at least $2^{\aleph_0}$ convergent sequences; there are no more than $2^{\aleph_0}$ real sequences... so equality ensues.2012-10-22
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    While you're answer is certainly correct, I don't think you can claim that *any* sequence must have $2^{\aleph_0}$ subsequences. Take the sequence $a,a,a,a,a...$ for example.2012-10-22
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    @Martian, perhaps this is why Asaf referred to "injective sequences".2012-10-22
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    @Gerry, this is an edit I made after MartianInvader's comment (as the timestamps would reveal).2012-10-22
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    @AsafKaragila, why are the set of subsequences has this cardinality?2012-10-22
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    @Martian, my apologies, I didn't think to look at the temporal sequence. Asaf, thanks.2012-10-22
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    @Tomás: A sequence is indexed by $\mathbb N$, so it has $2^{\aleph_0}$ distinct infinite subsets. Each subset defines a unique subsequence (uniqueness follow because we assumed the original sequence is injective, no point appears twice).2012-10-22
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    Take any sequence $(a_n)_{n \geq 1}$ which converges to $a$ and pick $a_0 \in \mathbb{R}$ arbitrarily. Then the $2^{\aleph_0}$ distinct sequences $(a_n)_{n\geq 0}$ all converge to $a$.2012-10-22
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    To make commenter's point even more explicit, for what it's worth, the set $\{(b,a,a,a,\ldots):b\in\mathbb R\}$ already gives the lower bound of $2^{\aleph_0}$.2012-12-30