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Decide if the set of the convergent sequences is a subspaces of the vector space of the infinite sequences $(a_i)^{\infty}_0$.

My solution:

For the convergence we have to satisfy following condition: $\lim_{n\to \infty} a_n = A$, so every sequence that belongs to the set of the convergent sequences must have a finite limit.

1) $\lim_{n\to \infty} (a_n + b_n )= \lim_{n\to \infty} a_n + \lim_{n\to \infty} b_n = A + B$ and that is a finite limit.

2) For $r\in \mathbb R$, $\lim_{n\to \infty} (ra_n) = rA$, the limit exists, so it is ok.

Sorry for my english.

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    If $r\in \mathbb R$, then $r$ is not infinite. And if the limit is $rA$, then the limit exists, so the sequence is convergent. – 2017-02-06
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    Real numbers are an infinite set, uncountable set, isn't it right? So why $r$ can not be infinite? – 2017-02-06
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    There are uncountably many real numbers. All of them are finite. Infinity is not a number. Except for that confusion, your answer is correct. – 2017-02-06
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    @Astrid There is no "is finite" or "is infinite" for a real number. The real numbers are all expressible as decimal expansions, and none of them could be considered "infinite." You can talk about the extended real numbers which adjoin elements that act like inifinity, and then talk about convergence to the adjoined elements, however. – 2017-02-06
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    Ok, so infinite can be a set of numbers but not a number itself. But I am little confused now, is it a subspace or not? If $r$ is finite, then it is convergent. – 2017-02-06
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    Your arguments are essentially correct, except the idea that "for really large $r$, the sequence $(ra_i)_i$ does not converge", as the other comments have pointed out. To make this a correct proof, you now have to quote the appropriate limit theorems about sequences. – 2017-02-06
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    Well, I can view $r$ as some fixed number from $\Bbb R$ and because our limit exists, it is a convergent sequence as was said above. – 2017-02-06

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