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This question already has an answer here:

So define the set of finite sequences to be $S={a_1,a_2,\cdots}$ where $a_k$ are in real numbers and only finitely many of them are non-zero. The set of infinite sequences is defined similarly except that we can have infinitely many non-zero terms. How do I prove that there does not exist a bijection between these two sets?

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marked as duplicate by Asaf Karagila cardinals Feb 6 '17 at 11:14

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  • $\begingroup$ As you tagged this as linear algebra, could it be that instead of bijection you mean isomorphism of real vector spaces? $\endgroup$ – Hagen von Eitzen Feb 6 '17 at 10:29
  • $\begingroup$ I was wondering how I show that the set of infinite sequences with entries in R is uncountably infinite- dimensional as a vector space over R. $\endgroup$ – John von Neumann Feb 6 '17 at 10:31
  • $\begingroup$ @JohnvonNeumann Isn't the set of infinite sequences of real numbers manifestly a countably infinite-dimensional vector space over R? $\endgroup$ – spaceisdarkgreen Feb 6 '17 at 10:52
  • $\begingroup$ spaceisdarkgreen : not so manifestly though $\endgroup$ – Max Feb 6 '17 at 11:37
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You don't, because there exist such bijections. The first one is (essentially) $\displaystyle\bigcup_{n\in \mathbb{N}} \mathbb{R}^n$, which has the same cardinality as $\mathbb{R}$, and the second one is $\mathbb{R}^\mathbb{N}$, which also has the same cardinality as $\mathbb{R}$.

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