# A local base problem

The following question is an example(without proof) in my lecture notes. I'm trying to figure out the proof.

Consider the subset of $\mathbb{R}^2$:

$$X = \{(0,0)\}\cup\{(2^{-n},0):n \in \mathbb{N}\}\cup\{(2^{-n},2^{-m}):n，m \in \mathbb{N}\}.$$

The points different from $(0,0)$ are assigned to their natural beighborhoods. But at $(0,0)$, for every $n\in \mathbb{N}$ and every function $f:\mathbb{N}\rightarrow\mathbb{N}$, we set

$$B(f,n) = \{(0,0)\}\cup\{(2^{-m},0):m \geq n\}\cup\{(2^{-m},2^{-l}):n，m \geq n, l\geq f(m)\}$$, then set $\mathcal{B}_{(0,0)} = \{B(f,n)\}_{f,n}$

Claim:

1. the assignment is a valid assignment of local bases.
2. $(0,0)$ is in the closure of $A = \{(2^{-n},2^{-m}):n,m\in\mathbb{N}\}$ but no sequence from A converges to $(0,0)$.

Since it is discrete (apart from $(0,0)$), every point is open set, so the claim that natural neighborhoods of points not $(0,0)$ are local bases is trivial. But I dont know how to argue about $(0,0)$. furthermore, should I use the result from claim 1 to prove claim 2?

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What does $n\in\gt\mathbb N$ mean? Is that supposed to be just $n\in\mathbb N$? You also have $m\ge\gt n$ and an obvious typo -- please take into account that dozens or hundreds of people will be reading your post and take some basic care not to waste their time with such mistakes that simple proofreading would catch. – joriki Apr 23 '12 at 15:28

I assume that by natural neighborhoods you mean the open neighborhoods in the relative topology inherited from the usual topology of $\mathbb{R}^2$. In other words, the local basis at $x\in X,x\neq(0,0)$ is $\mathcal{B}_x=\{X\cap U:U\text{ is open in } \mathbb{R}^2,x\in U\}$.

In order to be a system of local bases for a topology of $X$ the set $\{\mathcal{B}_x:x\in X\}$ has to have the following three properties: \begin{align*} &(1)\quad \mathcal{B}_x\neq\emptyset\text{ for every }x\in X;\ x\in U\text{ for every } U\in\mathcal{B}_x, \\ &(2)\quad \text{For every }x\in X,\ U,V\in\mathcal{B}_x\text{ there is a } W\in\mathcal{B}_x\text{ with }W\subseteq U\cap V, \\ &(3)\quad \text{For every }x,y\in X,\ x\in U\in\mathcal{B}_y \text{ implies that } V\subseteq U\text{ for some }V\in\mathcal{B}_x. \end{align*}

The property $(1)$ is clear for every $x\in X$.

The property $(2)$ is clear for every $x\in X\setminus\{(0,0)\}$. For $x=(0,0)$, take $U=B(f,n)$ and $V=B(g,m)$, then $B(\max\{f,g\},\max\{n,m\})\subseteq U\cap V$.

The property $(3)$ is clear if both $x$ and $y$ are in $X\setminus\{(0,0)\}$, so two cases remain. First if $x\in X\setminus\{(0,0)\}$, $y=(0,0)$ and $x\in B(f,n)$, then $B(f,n)\in\mathcal{B}_x$ so choose $V=B(f,n)$. The remaining case is $y\in X\setminus\{(0,0)\}$ and $x=(0,0)\in U\in \mathcal{B}_y$. Note that since $U$ is open in the relative topology, there is an $r>0$ so that $S=\{(x_1,x_2)\in X: 0\leq x_1\leq r, 0\leq x_2\leq r\}\subseteq U$. Now if you take $n\in\mathbb{N}$ with $2^{-n}\leq r$, then $B(\operatorname{id}_\mathbb{N},n)\subseteq S\subseteq U$.

We have shown that $\mathcal{B}_x$'s are a system of local bases for a topology. Note that while all the points of the form $(2^{-n},2^{-m})$ are isolated, the points $(2^{-n},0),n\in\mathbb{N}$ are not: for any neighborhood $U$ of the point $(2^{-n},0)$ there is an $m\in\mathbb{N}$ so that $\{(2^{-n},2^{-l}):l\geq m\}\subseteq U$.

Since $(2^{-n},2^{-f(n)})\in A\cap B(f,n)$ for any $f,n$, the point $(0,0)$ is in the closure of $A$. Assume that $(x_n)$ is a sequence in $A$. Write $x_n=(2^{-f(n)},2^{-g(n)})$. Then $B(g+1,1)$ is a neighborhood of $(0,0)$ containing no elements of $(x_n)$ so $(x_n)$ does not converge to $(0,0)$.

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