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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 $S$?


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up vote 13 down vote accepted

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? – Tomás Oct 22 '12 at 21:11
@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. – Asaf Karagila Oct 22 '12 at 21:11
@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). – Asaf Karagila Oct 22 '12 at 22:52
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 $a$. – commenter Oct 22 '12 at 23:08
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 $2^{\aleph_0}$. – Jonas Meyer Dec 30 '12 at 0:56

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