Difficulty with question about countable product of copies of $\mathbb{N}$ I had a lot of trouble trying to figure this problem out. Can someone explain how to properly do this?
Problem:Assume that $\mathbb{N}$ has the usual order. Let $\mathbb{N}^{\omega}$ denote the Cartesian product of a countable number of copies of the space $\mathbb{N}$. It can be endowed with the dictionary order in a natural way. Show that $\mathbb{N}^{\omega}$ with the dictionary order topology is uncountable, is not well ordered, and any set that does not have a least element does have a limit point.
Edit:I realized I showed a misunderstanding for the first part of the problem since for some reason I assumed each $s_{i,j} \in \{1,...,9\}$ instead of $\mathbb{N}$.
Attmempt:Suppose that $\{S_1,S_2,...\}$ is an enumeration of the elements of $\mathbb{N}^{\omega}$.Where $f:\mathbb{N} \rightarrow \mathbb{N}^{\omega}$ given by $f(n)=S_n$, is bijective.We will produce an element in $\mathbb{N}^{\omega}$ that is not on this list. Then there is no surjection from $\mathbb{N}$ onto $\mathbb{N}^{\omega}$, and our assumption that $f$ is bijective is false.
A list of elements in the enumeration is given by:
$$S_1=(s_{1,1},s_{1,2},...)$$
$$S_2=(s_{2,1},s_{2,2},...)$$
$$\vdots$$
$$S_n=(s_{n,1},s_{n,2},...)$$
$$\vdots$$
Where $s_{i,j} \in \mathbb{N}$ for all $i,j \in \mathbb{N}$.
Consider the element $A=(a_1,a_2,...) \in \mathbb{N}^{\omega}$ such that $a_k=7$ if $s_{k,k} \in \{1,2,3,4,5,6\}$, and $a_k=3$ if $s_{k,k} \in \{7,8,9\}$. $f$ is bijective so $A \in \mathbb{N}^{\omega}$ must appear on the list. However, the $k$'th component of $a_k$ of $A$ and $s_{k,k}$ of $S_k$ differ by between $4$ and $7$, so $A$ does not occur on this list and $\mathbb{N}^{\omega}$ is uncountable.
We will construct a set of elements that has no least element.  Let $S=\{S_1,S_2,...\}$ be the infinite set with elements given by:
$$S_1=(1,2,3,4,5,6,7,...)$$
$$S_2=(1,1,2,3,4,5,6,...)$$
$$S_3=(1,1,1,2,3,4,5,...)$$
$$S_4=(1,1,1,1,2,3,4,...)$$
$$S_5=(1,1,1,1,1,2,3,...)$$
$$\vdots$$
Then the succesor of every element in the set is less than it's predecessor in the lexicographical ordering. Since there is no smallest element in this nonempty subset of $\mathbb{N}^{\omega}$,$\mathbb{N}^{\omega}$ is not well ordered. Let $S$ be any subset of elements in $\mathbb{N}^{\omega}$ that does not have a limit point.
Note for the part below $(\downarrow$).I ran out of time, and had to turn in but knew it was wrong. I felt like it was wrong, because for a decreasing sequence of elements (in the dictionary order) must have reach a point where all the components are identical, up to a point where, the the components become eventually constant, to a point. However I had difficulty expressing this mathematically.
Any subset $S$ of $\mathbb{N}^{\omega}$ without a minimum contains a sequence of elements of the form $(a_1,a_2,a_3,...)$, where $a_i=a_j$ for $i \neq j$ holds for all terms after some index $t$ for all indices $n$ with $1\leq t \leq n<k$, and $a_k>a_n$. Consider the element in $\mathbb{N}^{\omega}$ having each entry the element $a_n \in \mathbb{N}$ after $a_t$ . Then any open set in $\mathbb{N}^{\omega}$ containing this element, contains elements in $S$. Since $S$ has no minimum there is an element $(a_1,a_2,...) \in S$ where $a_{t+1}=\dots=a_n$ for each natural number $n$, so that any open interval containing the element with all entries $a_n$ must contain an element of this form.
 A: Your diagonal argument for the first part is correct, though you don’t have to assume that $f$ is a bijection: you have in fact demonstrated that no function from $\Bbb N$ to $\Bbb N^\omega$ is a surjection, so in particular none of them is a bijection. In other words, there’s no need to make it a proof by contradiction.
Your construction of a set with no least element is also fine, though you could improve it slightly be giving an explicit definition of the sequences $S_n$: for $n,k\in\Bbb N$ let
$$s_{n,k}=\begin{cases}
1,&\text{if }k\le n\\
k-n+1,&\text{if }k>n\,,
\end{cases}$$
and let $S_n=\langle s_{n,k}:k\in\Bbb N\rangle$. You can even go on to argue that if $m<n$, then $s_{m,k}=1=s_{n,k}$ for $k\le m$, and $s_{m,m+1}=2>1=s_{n,m+1}$, so $S_n\prec S_m$, where $\prec$ is the lexicographic order on $\Bbb N^\omega$.
For the last part, suppose that $\mathscr{S}\subseteq\Bbb N^\omega$ has no least element. Let $\mathscr{S}_0=\mathscr{S}$. Given $\mathscr{S}_n\subseteq\mathscr{S}$ for some $n\ge 0$, let $a_{n+1}=\min\{\pi_{n+1}(S):S\in\mathscr{S}_n\}$, and let
$$\mathscr{S}_{n+1}=\{S\in\mathscr{S}_n:\pi_{n+1}(S)=a_{n+1}\}\,.$$
In this fashion we recursively construct a sequence $T=\langle a_n:n\in\Bbb N\rangle$; I claim that $T=\inf\mathscr{S}$ and hence that $T$ is a limit point of $\mathscr{S}$.
It’s straightforward to check that $T$ is a lower bound for $\mathscr{S}$, and I’ll leave that to you. It’s not much harder to check that if $T\prec T'$ for some $T'\in\Bbb N^\omega$, then there is an $S\in\mathscr{S}$ such that $S\prec T'$. Specifically, if $T\prec T'=\langle a_n':n\in\Bbb N\rangle$, let $m$ be minimal such that $a_m<a_m'$. Pick any $S\in\mathscr{S}_m$: $\pi_m(S)=a_m<a_m'$, and $\pi_k(S)=a_k=a_k'$ for $k<m$, so $S\prec T'$, as desired. Thus, $T'$ is not a lower bound for $\mathscr{S}$, and $T$ is therefore the greatest lower bound of $\mathscr{S}$. And in the process of showing this, we’ve also shown that $T$ is a limit point of $\mathscr{S}$: every open interval around $T$ contains an open interval $(T,T')_\prec$ for some $T'\succ T$, and we just showed that $(T,T')_\prec\cap\mathscr{S}\ne\varnothing$.
A: Suppose the sequence $S$ below of elements in $\mathbb{N}^{\omega}$ (dictionary order) has no least element. We’ll show that $S$ has a limit point.
$\begin{gather}
S_1=(s_{11},s_{12},\dots)\\
S_2=(s_{21},s_{22},\dots)\\
\vdots\\
S_n=(s_{n1},s_{n2},\dots)\\
\vdots\\
\end{gather}
$
Then it contains a strictly decreasing subsequence $T = S_{i_1},S_{i_2},\dots$, which can be constructed as follows.
Choose $i_1=1$. Now $S_{i_1}$ is not the least element of $S$, so there is a smallest index $i_2>i_1$ for which $S_{i_2}<S_{i_1}$. $\color{red}{\text{Note}}$ that by choosing $i_2$ as the smallest such index, $S_{i_2}$ is then smaller than every term preceding it in all of $S$.
But while $S_{i_2}$ is the least element of $S_1,S_2,\dots,S_{i_2}$, it is not the least element of $S$, because $S$ has no least element. Choose the smallest $i_3>i_2$ for which $S_{i_3}<S_{i_2}$. As $\color{red}{\text{noted}}$, the terms of $S$ smaller than $S_{i_2}$ do not precede $S_{i_2}$, so there is such an $i_3>i_2$.
Proceeding inductively, we have the following decreasing sequence with no least element, which is a subsequence of $S$.
$\begin{gather}
T_1=(t_{11},t_{12},\dots)\\
T_2=(t_{21},t_{22},\dots)\\
\vdots\\
T_n=(t_{n1},t_{n2},\dots)\\
\vdots\\
\end{gather}
$
It suffices to show that $T$ has a limit point in $\mathbb{N}^{\omega}$, because any limit point of $T$ is a limit point of its supersequence $S$.
The sequence $T$ is decreasing in dictionary order, so for each $i$, the sequence of integers in the $j$th positions, $t_{1j},t_{2j},\dots$, is decreasing for any $j$. The natural numbers are well-ordered, so $t_{1j},t_{2j},\dots$ is eventually constant and has a limit $t_j=\lim_{i\to\infty}t_{ij}$.
Then $\lim_{k\to\infty}T_k= (t_1,t_2,\dots)$.
