Find the closure of the set $[0,1] \times \{1\}$.

Question

In the space $[0,1] \times \{0,1\}$ with the dictionary order topology, find the closure of the set $[0,1] \times \{1\}$.

Can anyone please guide me through finding the closure of this set?

Answer 1

Of course $A= [0,1] \times \{1\}$ is part of the closure of $A$.

What points outside of $A$ have a neighbourhood that misses $A$? If $(x,t)$ has $0 < t < 1$, then the open interval $\langle(x,0),(x,1)\rangle$ contains $(x,t)$ and misses $A$, so these are not in the closure.

If $t=0$ we have the case $(0,0)$: this has a neighbourhood $[(0,0), (0,1)\rangle$ (as the minimum it has a special local base) that misses $A$, so it's also not in the closure of $A$.

If $x>0$, then a basic open neighbourhood of $(x,0)$ is an open interval $I=\langle(a,b), (c,d)\rangle$ with $(a,b) < (x,0)$ so that $a < x$ is forced by the definition of the order, (you see why the second coordinate $0$ creates different behaviour) and then $(a',1)$ is a point of $A$ that lies in that interval $I$, when $a'$ is chosen such that $a < a' < x$, which is possible. So every (basic) neighbourhood of $(x,0)$ with $x >0$ intersects $A$ and so lies in the closure of $A$.

Conclusion: $\overline{A}=((0,1] \times \{0,1\}) \cup \{(0,1)\}$ or equivalently $A \cup ((0,1] \times \{0\})$.