Closure of subbase elements

Question

Here is the question as stated:

Let $(X,\mathcal{T})$ be a nonempty topological space, and consider the set $C(X)$ of closed, nonempty sets in $X$. Set $\mathcal{T}^{*} = \mathcal{T} - \{ \emptyset\}$, the set of nonempty open sets in $X$. Set

$\mathcal{S} = \{S(U,V) | U, V \in \mathcal{T}^{*}\}$,

where for any $U,V \subseteq X$, we set

$S(U,V) = \{A \in C(X) | A \cap U \neq \emptyset, A \subseteq V \}$.

Since $S(X,X) = C(X)$, this set $\mathcal{S}$ is a subbase for a topology $C(\mathcal{T})$ on $C(X)$. We endow $C(X)$ with the topology $C(\mathcal{T})$.

I understand the setup, and I am trying to show the closure of $S(U,V)$ is $S(\overline{U},\overline{V})$.

I am also trying to show that $C(\mathcal{T})$ has a base $\mathcal{B}$ consisting of the sets

$B(U_{1},...,U_{n}) = \bigcap_{i=1}^{n} S(U_{i},\bigcup_{i=1}^{n} U_{i})$

where $U_{i} \in \mathcal{T}^{*}$ for $1 \leq i \leq n$ for some $n \geq 1$.

I think that for the second part I should show that the intersection of two sets of the form $B(U_{1},...,U_{n})$ have the same form but am not sure how to finish either proof, any help is welcomed. Thanks for your time.

Answer 1

You have the right idea for the second part, and it might be easier to do that before doing the first part: it’s generally easier to work with a base than with a subbase. You do have a notational error when you define the basic open sets. $B(U_1,\ldots,U_n$ cannot be

$$\bigcap_{i=1}^nS\left(U_i,\bigcup_{i=1}^nU_i\right)\;,$$

because you cannot use the same index variable for the intersection and for the union inside it: you mean that

$$B(U_1,\ldots,U_n)=\bigcap_{k=1}^nS\left(U_k,\bigcup_{i=1}^nU_i\right)\;,$$

so that each $U_k$ is paired up with the union $U_1\cup\ldots\cup U_n$.

Consider the basic open sets $B(U_1,\ldots,U_m)$ and $B(V_1,\ldots,V_n)$. Let $U=\bigcup_{k=1}^mU_k$, $V=\bigcup_{k=1}^nV_k$, and $W=U\cap V$. For $k=1,\ldots,m+n$ let

$$W_k=\begin{cases} U_k\cap W,&\text{if }1\le k\le m\\ V_{k-m}\cap W,&\text{if }m+1\le k\le n\;. \end{cases}$$

Now verify that

$$\begin{align*} B(U_1,\ldots,U_m)\cap B(V_1,\ldots, V_n)&=\bigcap_{k=1}^mS(U_k,U)\cap\bigcap_{k=1}^nS(V_k,V)\\ &\overset{*}=\bigcap_{k=1}^{m+n}S(W_k,W)\\ &=B(W_1,\ldots,W_{m+n})\;; \end{align*}$$

the starred step is the only one that requires a bit of work, and it’s not bad.

Answer 2

A common notation used for this topology on $C(X)$ ($2^X$ or $H(X)$ for hyperspace, or $\operatorname{exp}(X)$ are more common, as $C(X)$ is also used for the space of continuous functions on $X$..), the so-called Vietoris topology is to define for $A \subseteq X$

$$\langle A \rangle = \{F \in C(X)\mid F \subseteq A\} \text{ and } [A]=\{F \in C(X) \mid A\cap F \neq \emptyset\}$$

The most standard subbase for a topology on $C(X)$ is then defined by

$$\mathcal{S}=\{\langle U \rangle , [U] \mid U \neq \emptyset, U \text{ open }\}$$

Your sets $S(U,V)$ are then open in this topology as $\langle V \rangle \cap [U]$ and the sets in $\mathcal{S}$ are open in your topology as $[U] = S(U,X)$ and $\langle U \rangle = S(U,U)$. So both families of sets generate the same (Vietoris) topology.

The sets $B(U_1, \ldots, U_n)= \bigcap_{i=1}^n [U_i] \cap \langle \bigcup_{i=1}^n U_i \rangle$ are indeed part of the base generated by $\mathcal{S}$ too. These sets are also denoted by $\langle U_1, \ldots U_n \rangle$ in the literature. (for one subset $U_1$ there is no ambiguity with the earlier notation.)

In fact, if we consider a finite intersection of subbasic elements we can note that $\langle U \rangle \cap \langle V \rangle = \langle U \cap V \rangle$ so that one type is closed under finite intersections, so we can collect them together into one subbasic set.

So we can write $$\bigcap_{i=1}^m \langle U_i \rangle \cap \bigcap_{j=1}^n [V_j] = \langle (\bigcap_{i=1}^m U_i) \cap V_1, \ldots, (\bigcap_{i=1}^m U_i) \cap V_n , \bigcap_{i=1}^m U_i \rangle$$ so that the base is part of the standard base generated by the subbase. Hence the base generates the same topology as (either) subbase. Using "my" $\mathcal{S}$ is slightly easier for a proof that $X$ compact implies $C(X)$ compact using the Alexander subbase lemma (see my proof here), so that's why I prefer it.

Also, $[A]$ and $\langle A \rangle$ are closed when $A$ is closed, as $[A]=C(X)\setminus \langle (X\setminus A) \rangle$ and $\langle A \rangle = C(X)\setminus [X\setminus A]$, and it follows that $\langle C_1, \ldots C_n \rangle$ is closed when all $C_i$ are. This makes the inclusion

$$\overline{\langle U_1, \ldots U_n \rangle} \subseteq \langle \overline{U_1}, \ldots \overline{U_n} \rangle$$ quite clear.

The reverse inclusion need not hold in all spaces: let $X=\{0,1\}$ be Sierpiński space with $\{0\}$ open. Then $C(X)= \{\{1\},X\}$ and $\langle \{0\} \rangle = \emptyset$ while $\langle \overline{\{0\}}\rangle = \langle X\rangle = C(X)$. It does hold in $T_1$ spaces (so when all finite subsets are closed), for that I'll refer you this proof on this site.