0
$\begingroup$

Suppose that we have a family $\mathcal{F}=\left\{F_{i}\right\}_{i\in I}$ of subsets of X and $\tau(\mathcal{F})=\cap \left\{T_{i}:\mathcal{F}\subset T_{i}\right\}$ where $T_{i}$ are topologies that contain the family $\mathcal{F}$. We also know that

$\left\{\cap_{i=1}^{n}F_{i},\cup\cap_{i=1}^{n}F_{i},X,\varnothing \right\}\in \tau(\mathcal{F})$.

My problem is that in my notes, I have that $\mathcal{\beta}=\{\bigcup_{i\in I}(\bigcap_{j=1}^{n}F_{i_j})\mid F_{i_j}\in\mathcal{F}\}\cup \{X\}$ is a basis for the topology $\tau(\mathcal{F})$. But I can't really understand that. Because, if $\beta$ is a basis for the topology, then why we don't assume that $X$ is a basis for the topology.

Also is there a way to make $\mathcal{F}\cup \{X\}$ a base for the topology $\tau(\mathcal{F})$?

Any help would be great!

$\endgroup$
  • 1
    $\begingroup$ There are many set theoretic errors in your question. You have to be much, much more careful with set theory notation or your question is likely to be closed. Just as an example, you ask whether there is a way to make $\mathcal{F} \cup X$ a base for a topology on $X$. The quick answer is no, because each element of a base for a topology on $X$ is required to be a subset of $X$, but $\mathcal{F} \cup X$ generally contains elements that are not subsets of $X$, namely the elements of $X$ itself. I suspect, though, that this is not what you meant to ask. $\endgroup$ – Lee Mosher Dec 31 '17 at 17:03
2
$\begingroup$

First, let's clear up some misconceptions. You ask

why we don't assume that $\{X\}$ is a basis for the topology?

If $X$ is any space, then the set $\{X\}$ is a basis for the trivial or indiscrete topology on $X$. Moreover, the only possible basis for the trivial topology is $\{X\}$. So the only way for $\{X\}$ to be a basis for the topology $\tau(\mathcal{F})$ would be if $\mathcal{F}$ is empty.

You also ask

Also is there a way to make $\mathcal{F}\cup\{X\}$ a base for the topology $\tau(\mathcal{F})$?

Since $\mathcal{F}$ is an arbitrary collection of subsets of $X$, it is possible that $\mathcal{F}\cup\{X\}$ is not a basis for any topology at all. In particular, this collection may fail to satisfy the following condition. If $B_1$ and $B_2$ are two elements of $\mathcal{F}\cup\{X\}$ such that $B_1\cap B_2$ is nonempty, then for each $x\in B_1\cap B_2$ there must exist another element $B_3$ of $\mathcal{F}\cup \{X\}$ such that $x \in B_3$ and $B_3\subset B_1\cap B_2$. In fact, the notion that we want is that of a subbasis. It turns out that $\mathcal{F}\cup\{X\}$ is a subbasis for the topology $\tau(\mathcal{F})$.

Now to show that $\beta$ is a basis for the topology $\tau(\mathcal{F})$. A subbasis is just a collection of subsets of $X$ whose union is all of $X$. The basis $\beta$ is the set of arbitrary unions of finite intersections of elements in $\mathcal{F}\cup \{X\}$. Hence $\beta$ is the basis for the topology generated by the subbasis $\mathcal{F}\cup \{X\}$.

Let $T$ be the topology generated by the basis $\beta$ (or equivalently by the subbasis $\mathcal{F}\cup\{X\}$). We want to show that $T=\tau(\mathcal{F})$. First take $U\in T$. Then $U$ is a union of a finite intersection of elements in $\mathcal{F}\cup \{X\}$. If $U=X$, then clearly $U\in T_i$ for each $i$. Otherwise, $U$ is a union of finite intersections of elements of $\mathcal{F}$. Thus $U$ has to be open in any topology containing $\mathcal{F}$. Since each $T_i$ contains $\mathcal{F}$, it follows that $U\in\tau(\mathcal{F})$.

Next take $U\in\tau(\mathcal{F})$. Since $T$ is a topology containing $\mathcal{F}$, it follows that $U\in T$. Therefore $T=\tau(F)$, as desired.

$\endgroup$
  • 1
    $\begingroup$ Wouldn’t $\{X\}$ be a basis for $\tau \left( \mathcal F \right)$ if $\mathcal F = \{ \emptyset , X \}$? $\endgroup$ – Theoretical Economist Dec 31 '17 at 19:08
  • $\begingroup$ You are correct, I made a mistake. In both cases the resulting topology is trivial. Thanks for pointing it out. $\endgroup$ – Adam Lowrance Dec 31 '17 at 19:13
2
$\begingroup$

Given a set $X$ and a family of subsets $\mathcal{F} \subseteq \mathcal{P}(X)$. It is not necessary that $\mathcal{F}$ is a topology on $X$. What your book try to do is to $``$add$"$ as few subsets as possible to $\mathcal{F}$ to $``$make$"$ it into a topology. More precisely, we define the smallest topology generated by $\mathcal{F}$ to be $\tau(\mathcal{F}) = \cap\{\tau_i \mid \mathcal{F} \subseteq \tau_i \text{ and } \tau_i \text{ is a topology on } $X$\}$. You can check that $\tau(\mathcal{F})$ is a topology on $X$.

The next thing we want to do is to find a base (aka basis) for $\tau(\mathcal{F})$. Recall the definition of base: Given a topology $\tau_1$ on $X$. We say a family of subsets $\mathcal{B} \subseteq \mathcal{P}(X)$ is a base for the topology $\tau_1$ on $X$ if for every open set $U \in \tau_1$ we can find a family of basic open sets $\{B_j\} \subseteq \mathcal{B}$ such that $U = \cup \{B_j\}$.

In the question, you suggest we can take $X$ as a base for $\tau(\mathcal{F})$ on $X$. No, we can't, because $\mathcal{B} \subseteq \mathcal{P}(X)$ is a family of subsets but $X$ is not. A family of subsets would be $\{X\}$, not $X$. Now even if we consider $\{X\}$, it is still not a base for the topology $\tau(\mathcal{F})$ on $X$. Remember $\tau(\mathcal{F})$ is the smallest topology generated by $\mathcal{F}$. Therefore, we must have $\mathcal{F} \subseteq \tau(\mathcal{F})$, but in general $\mathcal{F} \nsubseteq\{X\}$. For example, we may take $X = \mathbb{R}$ and $\mathcal{F} = \{[0, 1]\}$.

Similarly, your second suggestion $\mathcal{F} \cup \{X\}$ is also in general not a base for $\tau(\mathcal{F})$ on $X$, because it has problem with finite intersections. Consider $X = \mathbb{R}$ and $\mathcal{F} = \{[-1, 0], [0, 1]\}$ and $\mathcal{S} = \mathcal{F} \cup \{X\} = \{[-1, 0], [0, 1], \mathbb{R}\}$. We claim $S$ is not a base for $\tau(\mathcal{F})$ on $X$. Since $\tau(\mathcal{F})$ is the smallest topology generated by $\mathcal{F}$, we have $\mathcal{F} \subseteq \tau(\mathcal{F})$. Also, $[-1, 0], [0, 1] \in \mathcal{F}$, so $[-1, 0], [0, 1] \in \tau(\mathcal{F})$ are open. By the axiom of topology, $\{0\} = [-1, 0] \cap [0, 1]$ is open. But $\{0\} \ne \cup\{S_j\}$ for any family of subsets $\{S_j\} \subseteq \mathcal{S}$, since $\cup\{S_j\}$ must be empty or contain $[-1, 0]$ or $[0, 1]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.