I am struggling to figure out how to prove this result, or if my method is correct. Consider the set $[0,1] \times [0,1]$ equipped with the dictionary order. $(a,b)<(c,d)$ if and only if $a<c$ or $a=c$ and $b<d$ in the order topology. Prove $(0,1)$ is a limit point of $\{(\frac{1}{n},0)|n \in \mathbb{N}\}$. This was an excercise so probably not requiring a proof, however I am curious as to how one would attempt to prove this.
Attempt since $\frac{1}{n} \rightarrow 0$, for any $\epsilon>0$, there is a natural number $N$, such that $n \geq N \implies \frac{1}{n}<\epsilon$, then for each $n \geq N$, the interval $((0,a),(\frac{1}{n},b)),a<1,b>0$, is open, nonempty and contains $(0,1)$ so this point is a limit point.
Also I wrote the $\epsilon$ based on my knowledge of convergent sequences however, I noticed this does not add anything in the proof, so I think this is wrong.
Best Answer
I will write $\preceq$ for the lexicographic order on $X=[0,1]\times[0,1]$. Let $U$ be any open nbhd of $\langle 0,1\rangle$. Then there $p=\langle a,b\rangle,q=\langle c,d\rangle\in X$ such that
$$\langle 0,1\rangle\in(p,q)_\preceq\subseteq U\,,$$
where $(p,q)_\preceq$ denotes the open interval between $p$ and $q$ in the lexicographic order. Then $\langle a,b\rangle\prec\langle 0,1\rangle\prec\langle c,d\rangle$, so $a=0$, $0\le b<1$, and $0<c\le 1$. There is an $m\in\Bbb Z^+$ such that $\frac1m<c$, and clearly $\left\langle\frac1n,0\right\rangle\in(p,q)_\preceq\subseteq U$ for each $n\ge m$. Thus, not only is $\langle 0,1\rangle$ a limit point of the set $\left\{\left\langle\frac1n,0\right\rangle:n\in\Bbb Z^+\right\}$, it is the limit of the sequence $\left\langle\left\langle\frac1n,0\right\rangle:n\in\Bbb Z^+\right\rangle$.