Let $X=[0,1]\times [0,1]$. We denote the set $ X $ with the subspace topology of $\mathbb{R}\times \mathbb{R}$ endowed with the topology order using the dictionary order.

(A) Find the closure in $ X $ of $A=\left \{ t\times 1/2 : 1/2\leq t<1\right \} $

(B) Find the interior in $ X $ of $B=\left \{ s\times t: 0\leq s\leq1, 1/2\leq t\leq3/4\right \} $

(A) I have thought a lot about this and I have come to the conclusion that $\bar{A}=A$ since $\bar{A}=A\cup {A}'$ and in this case I do not know if there is limit point, are there any limit point? How can I find them? The $ (0,1) $ would not be a limit point?

(B) int$(B)=\left \{ s\times t: 0< s<1, 1/2< t<3/4\right \} $ ? Thank you very much for your help.


First figure out what the lexicographic order topology on the plane is!

What does an open interval look like? This should give you a basis to answer the subspace question.

Added: you saw that the open sets of the plane are of the form $\{x\} \times (a,b)$ for all $x \in \mathbb{R}$ and all $a < b$ in the reals.

$B$ is indeed correct as can be seen this way.

$A$ is indeed closed: show its complement is open. The complement is just a union of (relatively open) $\{t\} \times (-1,1) \cap ([0,1] \times [0,1]$ for $t < \frac{1}{2}$ and $t=1$) and relatively open $\{t\} \times (-1,\frac{1}{2}) \cap ([0,1] \times [0,1])$ and $\{t\} \times (\frac{1}{2}, 1) \cap ([0,1] \times [0,1]0$ for $\frac{1}{2} \le t < 1$.

  • $\begingroup$ The topology dictionary order is generated by the base $\left \{ (a\times b , a\times c) : a<b, a<c, a,b,c \in \mathbb{R} \right \}$ $\endgroup$ – user425181 Aug 28 '17 at 19:57
  • $\begingroup$ Yes, but then what I say is right or wrong? Is that the only limit? Is that the int of B? $\endgroup$ – user425181 Aug 28 '17 at 21:03
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    $\begingroup$ @user425181 $a < b$ is not necessary, just $b < c$. We just have $\{a\} \times (b,c)$ as the interval $((a,b) ,(a,c))$. So open sets in this plane can be decided per $x$-coordinate. Now intersect these with $[0,1] \times [0,1]$ to get the subspace topology. $\endgroup$ – Henno Brandsma Aug 28 '17 at 21:05
  • $\begingroup$ When intersecting I get conditions $0\leq a,b,c \leq 1$ $\endgroup$ – user425181 Aug 28 '17 at 21:11
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    $\begingroup$ @user425181 $B$ is correct, and so is $A$, see edits. $\endgroup$ – Henno Brandsma Aug 28 '17 at 21:17

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