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May you could help me finding the simple $\tau$-closures of the subsets $A=\mathbb Q\times \{1\}$ and $B=Q\times \{0\}$

Backround: We consider a Moore plane $\Gamma:=\{(x,y)\in\mathbb R^2: y\ge 0\}$ with a topology $\tau$ through two neighborhood bases: Let $p=(x,y)$ and

$\mathcal{B}(p):=\{U_\epsilon(p) | 0<\epsilon\le y\}$ for y>0 and

$\mathcal{B}(p):=\{A_\epsilon(p) | \epsilon >0\}$ for $p=(x,0)$ and $A_\epsilon(p)=U_\epsilon((x,\epsilon))\cup\{(x,0)\}$

I do not know how to get the closures now.

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What does $U_\epsilon(p)$ denote? – Cameron Buie Apr 15 '13 at 22:09
$U_\epsilon(x):=\{y\in A| d(x,y)<\epsilon\}$ for $(A,d)$ a metric space and $x\in A, \epsilon>0$ – Voyage Apr 15 '13 at 22:12
up vote 1 down vote accepted

I will assume here that $U_\epsilon(p)$ is the open ball about $p$ of radius $\epsilon$ in the standard Euclidean metric on $\Bbb R^2$. Note that if we consider $H:=\{(x,y)\in\Bbb R^2:y>0\}$ as a subspace of $\Gamma$, the same subspace topology is induced as if we considered it as a subspace of $\Bbb R^2$ in the usual topology. Since $A$ lies entirely within $H$, then the closure of $A$ in $\Gamma$ will be the same as its closure in the plane under the usual topology. However, $B$ is closed in $\Gamma$. Indeed, taking any point of $\Gamma$ not on the $x$-axis, there is a neighborhood about that point that misses the $x$-axis, and so misses $B$. Thus, the only possible limit points of $B$ are on the $x$-axis. On the other hand, taking any $p$ on the $x$-axis and taking any $U\in \mathcal{B}(p)$, we have that the intersection of $U$ and the $x$-axis is precisely $\{p\},$ so $p$ isn't a limit point of $B$. Since $B$ has no limit points, then it is closed.

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Thanks or your answer, everything is clear excpect for "the closure of A in Γ will be the same as its closure in the plane under the usual topology", what do you mean with this sentence using math symbols. – Voyage Apr 15 '13 at 22:23
I mean that if $\tau_m$ is the topology on $\Bbb R^2$ induced by the open balls of the standard Euclidearn metric, then the $\tau$-closure of $A$ is the same as the $\tau_m$-closure of $A$--in particular, it is $\Bbb R\times\{1\}$. – Cameron Buie Apr 15 '13 at 22:26

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