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Let $X$ be an infinite set and $x_{0}$ a point of $X$. Let $B = \{\{x\} : x \neq x_0\} \cup \{A \subseteq X : x_0 \in A$ and $X - A$ is finite$\}$. Show that $B$ generates a topology over $X$.

For $B$ to be a base I must take a set of $X$ and there must be a set of $B$ such that given $U$ in $\tau$ and $S \in B$, we can write $U= \cup S$. My difficulty is to know which subset I take from $X$. Why use base properties?

Adam Zalcman
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Isadora Suhadolnick
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  • There is "a base for some topology" (even when no topology is given in advance), and a "base for a given topology $\tau$." These are different notions, and in this question you need to use the former. You posted three slightly different versions of this same question, and I posted a more specific comment as to how to show $B$ is a base (for some topology which is generated by $B$). Do you use a book, could you locate the two different definitions of a base, and try to appreciate the difference? Again, I think the question should be stated as: Show that $B$ is a base (for some topology). – Mirko Dec 26 '20 at 03:35
  • What I say in the previous comment is explained in detail in the following question and its answers. https://math.stackexchange.com/q/3483379 Please read that and report/ask for details and clarification. I could answer the three questions that you posted an hour ago or so (if someone else does not answer them before me), but I would like to make you first try to get some answers on your own. – Mirko Dec 26 '20 at 03:45

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First off, every family $S$ of subsets of some non-empty set $X$ generates a topology on $X$ (if $S$ is used as a so-called subbase). So, there is nothing to show, the way the question is stated.

Note that a topology on $X$ is not given in your question. You cannot say "given $U$ in $\tau$", because there is no $\tau$ here, yet. There is only $B$, for the moment, and $\tau$ will come later, it will be the topology generated by the base $B$.

You would want to define $\tau$ as follows. A set $U$ is going to be open, if for every $x\in U$ there is $B_x\in B$ such that $x\in B_x\subseteq U$. (Equivalently, there is $B_U\subseteq B$ such that $U=\cup B_U=\cup\{C:C\in B_U\}$.)

But, if you define $\tau$ as above, it is not necessarily the case (in general) that $\tau$ will be a topology, because in order for $\tau$ to be a topology it must satisfy certain conditions:
(t1) $X$ and $\varnothing$ are in $\tau$ (i.e. $X$ and $\varnothing$ are open),
(t2) the union of any subfamily of $\tau$ is a member of $\tau$ (i.e. the union of any family of open sets is open), and
(t3) the intersection of any finite subfamily of $\tau$ is a member of $\tau$ (i.e. the intersection of any finite family of open sets is open).

So, if you start with $B$ and define $\tau$ as above, in order for $\tau$ to satisfy conditions (t1), (t2), (t3), the family $B$ itself must satisfy certain conditions. These are:
(b1) $\cup B=X$ (also written as $\cup\{C:C\in B\}=X$), and
(b2) If $B_0,B_1\in B$ and if $x\in B_0\cap B_1$ then there must be $B_2\in B$ such that $x\in B_2\subseteq B_0\cap B_1$.

Before you could talk about $\tau$ you need to verify that the family $B$ satisfies conditions (b1) and (b2) above. Consider this part of my answer so far to be a hint (that is, I am just trying to clarify to you your question, since I believe I know what you are asking, and you didn't state your question the way one would expect), and please try to verify (b1) and (b2) on your own. Report your progress, and ask more questions (you could use comments for your question or write comments to my answer for that), and I may provide the rest of the details, if you don't get them, but it would be better for you if you try to get the details on your own.

Here are the details. Condition (b1) is easy to verify (e.g. use that $X\in B$ since $X\setminus X=\varnothing$ which is finite). In order to verify (b2) take any nonempty $A_0,A_1\in B$ and consider cases. If $x_0\in A_0\cap A_1$ then $X\setminus A_0=F_0$ and $X\setminus A_1=F_1$, where each of $F_0$ and $F_1$ is finite, and hence the set $F=F_0\cup F_1$ is finite (and $x_0\not\in F$). Hence $A_0\cap A_1=(X\setminus F_0)\cap(X\setminus F_1)=X\setminus(F_0\cup F_1)=X\setminus F\in B$ (so we are done with this case, and it was not necessary to consider specific elements of $A_0\cap A_1$).
If $x_0\not\in A_0$ then it must be that $A_0=\{x\}$ for some $x\not=x_0$. Then either $A_0\cap A_1=\{x\}\in B$ or $A_0\cap A_1=\varnothing$ (in which case there is noting to prove). The case when $x_0\not\in A_1$ is similar.

Mirko
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