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Is the set: $$S=\{\text{Sequences of functions } \mathbb N\to\mathbb R\}$$

countable or uncountable

I think the set of functions from $\mathbb N \to \mathbb R$ is uncountable and so this should also be uncountable. But I cant understand what is meant by a "sequence" here.

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    $\begingroup$ Can you point us to the source, or is this verbatim? It seems unclear to me as well. $\endgroup$ – Jonathan Y. Sep 26 '13 at 14:11
  • $\begingroup$ @JonathanY. This is quiz problem which was asked and there was a question for set of functions from N to R . I was not able to distinguish how they were different. $\endgroup$ – user2179293 Sep 26 '13 at 14:17
  • $\begingroup$ It would help if you could give us an exact quote, or a link to the question. As is, what you're asking isn't clear. $\endgroup$ – Jonathan Y. Sep 26 '13 at 14:20
  • $\begingroup$ @JonathanY. S = {Sequences of functions from N to R.} This is verbatim $\endgroup$ – user2179293 Sep 26 '13 at 14:28
  • $\begingroup$ If that means $(\mathbb{R}^\mathbb{N})^\mathbb{N}$, then you're right, it's uncountable. $\endgroup$ – Jonathan Y. Sep 26 '13 at 14:28
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I'm assuming you mean the set of all sequences with values in $\mathbb{R}$; that is the set of all $(a_n)_{n\in\mathbb{N}}$ such that $a_n\in\mathbb{R}$. This set is uncountable, for the following reason:

Take $\mathbb{R}\to S$ that sends $x$ to the constant sequence $(a_n)_{n\in\mathbb{N}}$, where $a_n=x$ for all $n$. This function is injective, and so the cardinality of $S$ must be greater than $\mathbb{R}$, which is uncountable.

Edit If $S$ is the set of sequences of functions from $\mathbb{N}$ to $\mathbb{R}$, that is, the set of $(f_n)_{n\in\mathbb{N}}$ where $f_n:\mathbb{N}\to\mathbb{R}$, then the proof is similar. We see that a sequence of real numbers can be seen as a sequence of constant functions from $\mathbb{N}$ to $\mathbb{R}$, and so the set I showed is uncountable above is contained in the set $S$, and therefore the set you are looking at is uncountable.

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  • $\begingroup$ This is the set of functions $\mathbb{N}\to\mathbb{R}$, which @user2179293 acknowledged as uncountable, hence I doubt he meant the same set. $\endgroup$ – Jonathan Y. Sep 26 '13 at 14:18
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Hints:

1) First prove that if there is a one-one map from $A\rightarrow B$ and $A$ is uncountable, then so is $B$.

2) Then prove that there is always a one-one map from $A$ to the set of sequences with values in $A$.

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