Is this set countable or uncountable?

Question

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.

Can you point us to the source, or is this verbatim? It seems unclear to me as well. — Jonathan Y., Sep 26 '13 at 14:11
@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. — user2179293, Sep 26 '13 at 14:17
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. — Jonathan Y., Sep 26 '13 at 14:20
@JonathanY. S = {Sequences of functions from N to R.} This is verbatim — user2179293, Sep 26 '13 at 14:28
If that means $(\mathbb{R}^\mathbb{N})^\mathbb{N}$, then you're right, it's uncountable. — Jonathan Y., Sep 26 '13 at 14:28

rfauffar · Answer 1 · 2013-09-26T14:28:35.087

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.

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

score 1 · Answer 2 · answered Sep 26 '13 at 14:53

1

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 $A$.

answered Sep 26 '13 at 14:53

WillO

3,141

Is this set countable or uncountable?

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