Proving the set of the strictly increasing sequences of natural numbers is not enumerable. How would one proceed to prove this statement?

The set of the strictly increasing sequences of natural numbers is not enumerable.

I've been trying to solve this for quite a while, however I don't even know where to start.
 A: For any infinite subset of the natural numbers, you can list its members in increasing order, and then you have a sequence that is strictly increasing.
Moreover, if you take two distinct infinite subsets of the natural numbers,
you get two different sequences.
(There is some number $k$ that is in one of the subsets and not the other,
and this number occurs in one sequence and not the other.)
So the number of strictly increasing sequences of natural numbers is at least as great as the number of infinite subsets of natural numbers.
A: As other answers note, there are lots of fancy ways to prove this. But we can always go back to the basics. A straightforward diagonalization proof-by-contradiction suffices. Suppose there is such an enumeration. Maybe this is it:
1 --> 1, 2, 3, 5, ...
2 --> 4, 5, 7, 100, ...
3 --> 1, 2, 3, 8, ...
4 --> 2, 4, 5, 6, ...

Now take the first number of sequence one, and add one to it.  That's our first number: 2.
Now take the second number of sequence two - 5 - and the number from the previous step - 2. Take the larger and add one: 6.
Now take the third number of sequence three - 3 - and the number from the previous step - 6. Take the larger and add one: 7.
Now take the fourth number of sequence four - 6 - and the number from the previous step - 7. Take the larger and add one: 8.
Keep doing that and construct the sequence of monotone increasing naturals:
2, 6, 7, 8, ...

By assumption, this sequence is in our enumeration, but where can it be? It cannot be at spot n for any n because by its construction the nth element of this sequence is larger than the element at spot n of the nth sequence.
That's a contradiction, and therefore there cannot be any such enumeration.
A: Given a sequence $(a_0, a_1, \ldots, )$, map it to the sequence $(a_1-a_0, a_2-a_1, \ldots)$.
The image of such map would be the sequence of all natural numbers . Indeed the map is surjective since given a sequence of natural number $(b_0, b_1, \ldots)$, we can find a preimage $(a_0, a_1, \ldots)$ which satisfy $a_0 =1$, $a_{i+1} = b_i + a_i , \forall i \geq 0$. That is the cardinality of the set of increasing sequence of natural number is at least as big as the set of sequence of natural number.
By Cantor diagonalization, we know the set of all sequence of natural number is not countable.
A: There are uncountably many subsets of $\Bbb N$, but only countably many finite subsets, hence uncountably many infinite subsets.
Every strictly increasing sequence of naturals corresponds to an infinite subset of $\Bbb N$.
A: Hint: Could you uniquely associate to a sequence of $1$s and $2$s a strictly increasing sequence of natural numbers?
A: We can define a very simple injection from the real numbers in the interval $[1,10)$ to your set by mapping $x \in [1,10)$ to the sequence $$\lfloor x \rfloor, \lfloor 10x \rfloor, \lfloor 100x \rfloor, \lfloor 1000x \rfloor, \dots.$$
For example, $\pi$ would map to the sequence $$3, 31, 314, 3141, 31415, 314159, \dots.$$
There is some straightforward checking to do that


*

*the resulting sequence is increasing, and

*two different real numbers map to different sequences.


Once we've done that, this argument shows that the strictly increasing sequences have at least the cardinality of the set $[1,10)$, which is continuum.
A: Map any strictly increasing sequence $(a_n)$ to a sequence $(b_n)$ of its increments modulo $2$:
$$\{0,1\}\ni b_n \equiv a_{n+1}-a_n \pmod 2$$
This is, of course, not bijective or even injective, but it is surjective mapping, hence the cardinality of the set of $a$ sequences is not less than that of $b$ sequences.
And the latter is known to be strictly greater than $|\mathbb N|$ because $b$ are binary sequences, which are one-to-one representation of $2^\mathbb N$. They can also be bijectively mapped onto $\mathbb R$.
A: For a countably infinite set $F$ of strictly increasing functions from $\Bbb N$ to $\Bbb N$ let $F=\{f_n:n\in \Bbb N\}.$ Define $g:\Bbb N \to \Bbb N$ by $g(n)=1+\sum_{j=1}^n f_j(n).$ 
Then $g(n+1)-g(n)=f_{n+1}(n+1)+\sum_{j=1}^n (f_j(n+1)-f_j(n))>0$ so $g$ is strictly increasing. 
And $g\not \in F$ because $g(n)\geq 1+f_n(n)>f_n(n)$, so $g\ne f_n$ for any $n.$
A: For $\alpha > 1$ consider the strictly increasing sequence $f_{\alpha}(n)= [n \alpha]$
The map $\alpha \mapsto f_{\alpha}(\cdot)$ is injective, since $\lim_{n\to \infty} \frac{[n\alpha]}{n} = \alpha$.
A: The map from $\{0,1\}^{\mathbb N}$ into the set of strictly increasing sequences of natural numbers given by
$$(a_n) \to (1+a_1,3+a_2, 5+a_3,7+a_4, \dots)$$
is injective. Since $\{0,1\}^{\mathbb N}$ is uncountable, we're done.
A: There are uncountable number of positive real numbers, take 
{An}=cn where c is a positive real number
Do this with every positive numbers it is increasing and uncountable.
A: The family of all strictly increasing sequences of natural numbers,  if regarded as a family of functions from naturals into naturals, is a dominating family under both pointwise and mod finite orders, so it size has to be larger than or equal to the dominating number $\mathfrak{d}$, which is a well-known uncountable cardinal invariant of the continuum.
A: Let $\{{s_i}_j\}$ be a countable list of strictly increasing sequences; define $\{c_i\}$ via $c_i = \max (c_{i-1}, {s_i}_i)+1$ and .... presto, Cantor!
But it  could be simpler (depending on one's idea of simple) to reduce to things we already know are uncountable.
For simplicity, we can consider the sequence of differences between terms and not have to worry about the terms being increasing. i.e $a_{i+1} > a_i$ so $b_{i+1} = a_{i+1} - a_i > 0$ so $b_{i+1} \in \mathbb N$ and if $b_0 = a_0$. We have a one-to-one correspondence between $\{$ all increasing sequence of natural numbers $\}$ and $\{$ all sequences of natural numbers $\}$.
$\{$ all sequences of natural numbers $\} \supset \{$ all sequences of 1.... 10$\} \cong \{$ all sequences of 0....9 $\} \cong \{$all real numbers between 0 and 1$\}$ which is uncountable.
