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Claim: Prove that the set of all increasing sequences $n_1

Here's my attempt:

Let $X=\{(x_n)_{n\in N};(x_n)_{n\in N}$ is increasing$\}$. Consider $f:N\rightarrow X$. We claim that no $f$ can be surjective.

Denote $(u_n)_{n\in N}$ the image of $f$ at the point $u\in N$, ie, $f(u) =(u_n)_{n\in N}$ . We construct a sequence $(s_n)_{n\in N}$ such that $(s_n)_{n\in N}$$\neq f(s)$ for all $s\in N$.

For, construct $(s_n)_{n\in N}$ taking $s_1=\prod_{k=1}^{\infty} f(k)_1$, $s_2=\prod_{k=1}^{\infty} f(k)_2$, ... , $s_n = \prod_{k=1}^{\infty} f(k)_n$, ...

By the monoticity of multiplication, we guarantee that $(s_n)_{n\in N} \in X$. But by the way we built it, it is such that $(s_n)_{n\in N} \neq f(s)$ for all s natural. Hence, $f$ cannot be surjective and therefore X is not countable.

Is it correct? And if yes, i find pretty intuitive that $(s_n)_{n\in N} \neq f(s)$ for all s natural. But can someone turn it out more rigorous and clear?

EDIT:

I came up with a new solution and better notation. Here it is:

Let $X=\{(x_n)_{n\in N};(x_n)_{n\in N}$ is increasing$\}$. Consider $f:N\rightarrow X$.

Denote $(u_n)_{n\in N}$ the image of $f$ at the point $u\in N$, ie, $f(u) =(u_n)_{n\in N}$ . We may write this sequence as $f(u)= (f(u)_1,...,f(u)_n,...)$. Also define $A_u=\{f(u)_1,...,f(u)_n,...\}$ as the set of all elements of the sequence $f(u)$. We define this set for every $u \in N$.

Now we construct a new sequence $(s_n)_{n\in N}$ inductively given by:

Define $f(s)_1 $= the smallest element of $A_1$.

Supposing defined $f(s)_1,...,f(s)_n$, let $S_{n+1} = \{x \in A_{n+1}; x>y$, where $y$ is the smallest element of $A_{n}\}$.

Then take $f(s)_{n+1}$= smallest element of $S_{n+1}$.

Therefore the sequence $(s_n)_{n\in N}$ is now defined inductively, and by construction $(s_n)_{n\in N}$ is increasing. Therefore, $(s_n)_{n\in N} \in X$ and $(s_n)_{n\in N}$ is such that $(s_n)_{n\in N} \neq f(s)$, for all $s \in N$. Therefore $f$ is not surjective and hence X is not countable.

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    you should first specify as to whether you mean sequences of natural numbers or those of real numbers. Secondly, please note that in defining the individual terms of the new sequence, you're multiplying an _infinite number_ of real numbers. – 2017-01-11
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    Increasing sequences of natural numbers. – 2017-01-11
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    By $s_1=\prod_{k=1}^{\infty} f(k)_1$ I understand the product of the first element of all sequences in $X$, am I right? Similar for $s_2, s_3, \ldots $. In that case that product may go to infinity. – 2017-01-11
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    Note that there's an obvious bijection between the set of all strictly increasing sequences of natural numbers and the set of all infinite sets of natural numbers. That could be useful, depending on what you already know. – 2017-01-11

2 Answers 2

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Here is another approach. Suppose the set is countable, hence we can make a list of its elements $f_1, f_2, f_3, \ldots$. Consider the function $f$ defined by $f(1)=f_1(1)+1$ and $$ f(n)=f_n(n)+f(n-1)+1$$ for $n>1$. Then $f$ is increasing, but $f$ differs from every $f_i$ of the list. Therefore, as we can't make a list of its elements, the set is uncountable.

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    Why not $f(n) = f_n(n) + 1$ for all $n?$ – 2017-01-11
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    We need $f$ to be increasing. Letting $f(1)=f_1(1)+1$ and $f(2)=f_2(2)+1$ we are not sure that $f_1(1) \leq f_2(2)$ – 2017-01-11
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    of course. thanks – 2017-01-11
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If you already know that there are uncountably many subsets of $\mathbb N$, then you can solve this problem by exhibiting a one-to-one map from subsets of $\mathbb N$ to increasing sequences of natural numbers, for example the map sending any $X\subseteq\mathbb N$ to the sequence $(x_n)$ where $x_n=2n$ for $n\in X$ and $x_n=2n+1$ for $n\notin X$.