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Let $X = \{0,1,2,3,\ldots\}$ (the non-negative integers), let $$B_1 = \{\{n\} : n \in X \text{ and }n > 0\}= \{\{1\}, \{2\}, \{3\},\ldots\}$$ $$B_2 = \{Z \subset X : X \setminus Z = \{1,2,\ldots n\} \text{ for some }n \in \mathbb{N} \}$$

a) Prove that $B$ is a basis for a topology on $X$.

b) Let $T$ be the topology from part 1. Prove that $(X; T )$ is $T_2$.

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Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove") to be rude when asking for help; please consider rewriting your post. –  Zev Chonoles Apr 22 '12 at 3:38
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Also, you said what $B_1$ and $B_2$ are, but not $B$. Perhaps $B=B_1\cup B_2$? –  Zev Chonoles Apr 22 '12 at 3:39
    
Forget my previous comment. I was first confused by T2 notation. Of course, you mean Hausdorff $T_2$-space. –  Vadim Apr 22 '12 at 4:26
    
The space given by $B=B_1\cup B_2$ is homeomorphic to $\{0\}\cup\{\frac1n; n\in\mathbb N$ as the subspace of the real line with the usual topology. (Perhaps this might help you visualize the problem.) Anyway, you just need to write down the definitions and check. –  Martin Sleziak Apr 22 '12 at 4:35
    
Zev Chonoles, sorry I'll edit my post and you are right B=B1 U B2 –  user24874 Apr 22 '12 at 13:18

2 Answers 2

To show that $\mathscr{B}$ is a base for a topology on $X$, you must show two things:

  1. For each $x\in X$ there is some $B\in\mathscr{B}$ such that $x\in B$, and
  2. for any $B_0,B_1\in\mathscr{B}$ and any $x\in B_0\cap B_1$, there is some $B\in\mathscr{B}$ such that $x\in B\subseteq B_0\cap B_1$.

In your case $\mathscr{B}=\mathscr{B}_1\cup\mathscr{B}_2$, where $$\mathscr{B}_1=\Big\{\{n\}:n\in X\text{ and }n>0\Big\}\;,$$ and $$\mathscr{B}_2=\{X\setminus F:F=\{1,\dots,n\}\text{ for some }n\in\Bbb N\}\;.$$

If $n\in X\setminus\{0\}$, then $n\in\{n\}\in\mathscr{B}_1\subseteq\mathscr{B}$, and $0\in X\setminus\{1\}\in\mathscr{B}_2\subseteq\mathscr{B}$, so condition (1) is satisfied.

Condition (2) is also easy to check. Suppose that $B_0,B_1\in\mathscr{B}$ and $n\in B_0\cap B_1$. If $B_0$ and $B_1$ both belong to $\mathscr{B}_1$, then clearly $B_0=B_1=\{n\}$, and in (2) we can take $B$ to be $B_1$. In fact, you can easily check that if either of $B_0$ and $B_1$ belongs to $\mathscr{B}_1$, we can take that set to be the $B$ of (2): if a member of $\mathscr{B}_1$ intersects a member of $\mathscr{B}$, it must already be a subset of that member. Thus, the only case that requires a little work is when $B_0,B_1\in\mathscr{B}_2$, and I leave you to check that in that case one of them is again a subset of the other.

To show that the resulting topology is Hausdorff ($T_2$), you need only show that if $m$ and $n$ are distinct elements of $X$, there are $B_m,B_n\in\mathscr{B}$ such that $m\in B_m$, $n\in B_n$, and $B_m\cap B_n=\varnothing$. This is dead easy of neither $m$ nor $n$ is $0$: just take $B_m=\{m\}$ and $B_n=\{n\}$. Separating $0$ and some $n>0$ is almost as easy: for you neighborhood of $n$ you will of course choose $\{n\}\in\mathscr{B}_1$, and for your neighborhood of $0$ you'll want some member of $\mathscr{B}_2$ that misses $n$. The simplest choice is $X\setminus\{1,\dots,n\}$, but any $X\setminus\{1,\dots,m\}$ with $m\ge n$ will work just as well.

You should try to convince yourself that this topology on $X$ makes $\langle 1,2,3,\dots\rangle$ a sequence converging to the point $0$. In fact, as Martin noted in the comments, $X$ looks just like the subset of $\Bbb R$ consisting of $0$ and the reciprocals $1/n$ for $0<n\in\Bbb N$. Call that subset $Y$. If you take as a base the sets $\{1/n\}$ for integers $n>0$ and the sets $Y\setminus\{1,\frac12,\dots,\frac1n\}$ for integers $n>0$, imitating $\mathscr{B}_1$ and $\mathscr{B}_2$ in your problem, you get the same topology on $Y$ that it inherits from the usual topology on $\Bbb R$, in which $\langle 1,\frac12,\frac13,\dots\rangle$ converges to $0$.

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I think you mean show $B_i$ is a basis for a topology on $X$, where $B_1$ and $B_2$ are as given. To show $B_1$ is a basis for a topology on $X$, you need only show that if $A,C \in B_1$ and have non-empty intersection, then there is a $V \in B_1$ such that $V$ is contained in $A$'s intersection with $C$. But with $B_1$ if $A \cap C$ is non-void, $A=C$, so you may choose $V=A=C$ in this case and establish that these sets in $B_1$ are a basis for a topology on $X$. It's a different but similar thing with $B_2$. If $A \in B_2$ and $C \in B_2$ then $X-A =\{1,2,3\ldots,n\}$ and $X-C = \{1,2,3,\ldots m\}$ for some $n$ and $m \in N$. Now $X-A \cup X-C =\{1,2,3,\ldots\max \{m,n\}\}$ so that $X-(A \cap C) = X \cap (X-A) \cup (X-C)$ and from the definition of $B_2$ we see $A \cap C$ is IN $B_2$ (this is stronger than what you need, but does imply that $B_2$ is a basis). I'll give you some time to think about what neighborhoods of points look like in both topologies to decide if $T$ induced by $B_1$ makes $X$ a $T_2$ space.

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I welcome you to the site. You may want to TeXify your posts. Although I don't understand the math of this posts, I did follow the math of some your previous posts. I liked some of them. But, I didn't upvote (neither did I downvote.) because they are not TeXified. This site allows you to have a much cleaned display of Math formulas. An inline formula is surrounded by $ while a display equation is surrounded by $$. Zev Chonoles has edited this post. You may want to see that for more details. With Sincere Regards, –  user21436 Apr 22 '12 at 4:02
    
James Auld, thank you so much for your help. –  user24874 Apr 22 '12 at 13:46

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