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Prove that the set of all finite sequences of real numbers has the same cardinality as the set $\mathbb{R}$ of reals.

I can not understand the purpose of the question.

Math1000
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clara
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  • See here. – OR. Apr 01 '15 at 16:11
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    Not really a duplicate -- a finite sequence is not the same as a finite subset, and a priori one could imagine there are more sequences than subsets (since the obvious map from sequences to subsets is surjective but not injective). – hmakholm left over Monica Jan 30 '16 at 01:59
  • I agree with @hmakholmleftoverMonica. Yet, there is a newer question that covers precisely this question (a bit more general) https://math.stackexchange.com/questions/3696710/cardinality-of-finite-sequences-of-infinite-set/3920798#3920798 – user1868607 Nov 24 '20 at 12:32

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Let $A\sim B$ mean $A,B$ have the same cardinality. You want to show $\cup_{n=1}^\infty\mathbb {R}^n \sim \mathbb {R}.$ Assuming you know that each $\mathbb {R}^n\sim \mathbb {R},$ we can do this: There are bijections from ${R}^n$ to $[n,n+1)$ for each $n$ (because $[n,n+1)\sim \mathbb {R}),$ hence there is a bijection from $\cup_{n=1}^\infty\mathbb {R}^n$ to $[1,\infty) \sim \mathbb {R}.$

zhw.
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