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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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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
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

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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