Limit Points of $\{(\frac{1}{2}, y) \mid 0Consider the following set in dictionary order topology on $[0,1] \times [0,1]$ $$E = \{(\frac{1}{2}, y) \mid 0<y<1\}$$  Are  ​$(\frac{1}{2}, 0)$ and $(\frac{1}{2}, 1)$  limit points of $E$?
I think yes.
My ATTEMPT:
First prove $(\frac{1}{2}, 0)$ is a limit point.
Let  $U : = ((a,b), (c,d))$ be a basic open set containing $(\frac{1}{2}, 0)$. Therefore, $(a,b) < (\frac{1}{2}, 0) < (c,d)$. Let  $a<\frac{1}{2}$, then if $c<\frac{1}{2}$, $U$ intersects $E$ and if $c = \frac{1}{2}$, $0<d$, $U$ also intersects $E$. Now consider $a = \frac{1}{2}$ and $b<0$
Now I can not proceed further. Please help me.
 A: If $(a,b) < (\frac12,0) < (c,d)$ we know that $\frac12 \le c$ and if $c=\frac12$, then $d>0$. This implies that always $(\frac12, d') \in E \cap \langle (a,b), (c,d) \rangle$, where $d' = \frac12$ if $c > \frac12$ and $d'=\frac{d}{2}$ if $c=\frac12$, as in either case $(\frac12,0) < (\frac12,d') < (c,d)$.
The $(\frac12,1)$ case is similar, but on the left side.
In fact $(\frac12,0) = \inf E$ and $(\frac12,1) = \sup E$ in this ordered set, but not in $E$, so it follows from that too that they're limit points. Also $\{\frac12\}\times [0,1]$ is order isomorphic to $[0,1]$ in the obvious way, and this is another way to see it.
A: Hint: If $A \subset (X,\tau)$, and $x\in X$, then we say that $x$ is a limit point of $A$ if every open set containing $x$ intersects $A\setminus \{x\}$.
What do open sets in this topology look like? They are of one of the following forms:

*

*$[0\times 0, x_1\times y_1)$

*$(x_1\times y_1, x_2\times y_2)$

*$(x_1\times y_1, 1\times 1]$
Consider an arbitrary open set containing $\left(\frac12, 0\right)$ and proceed. For $\left(\frac12, 1\right)$, the argument is similar.
A: @Henno Brandsma Sir, I am writing here that I have understood from your answer.
Let  $U : = ((a,b), (c,d))$ be a basic open set containing $(\frac{1}{2}, 0)$. Therefore, $(a,b) < (\frac{1}{2}, 0) < (c,d)$, which implies that $a < \frac{1}{2}$ and  $ \frac{1}{2} \leq c $ (and if $c = \frac{1}{2}$, then $d>0$). Thus, $((a,b), (c,d)) \cap E \neq \emptyset$ and hence $(\frac{1}{2}, 0) \in \overline{E}$.
Similarly, let  $U : = ((a,b), (c,d))$ be a basic open set containing $(\frac{1}{2}, 1)$. Therefore, $ a \leq \frac{1}{2}$ (and if  $a = \frac{1}{2}$, then $b< 1$) and $\frac{1}{2} < c$. Therefore, $((a,b), (c,d)) \cap E \neq \emptyset$, and hence $(\frac{1}{2}, 1) \in \overline{E}$.
