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Consider $(X, \tau)$ where $$\tau = \{\emptyset, X, \{a \}, \{c\}, \{a,c \},\{a,b \},\{b,c \} \},$$

and the nonbasis $$B = \{ \{a\}, \{c\},\{a,b\},\{ b,c\} \}.$$

My book says $\emptyset$ can be generated by an "empty union of memebers in $B$". I take it they mean $$\emptyset = \cup \emptyset.$$

But how can this happen when $\emptyset \notin B$?

Lemon
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1 Answers1

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A subset $S \subseteq B$, is itself a set of subsets of the topological space $X$; then $\bigcup S$ is the subset of $X$ consisting of those elements which lie in some set in $S$.

More precisely, if $S \subseteq \mathcal{P}(X)$ then $$\bigcup S = \{ x \in X : x \in U\ \text{for some}\ U \in S \}$$

Now $\varnothing \subseteq S$ since the empty set is the subset of every set (so it's definitely a subset of $S$). And $$\bigcup \varnothing = \{ x \in X : x \in U\ \text{for some}\ U \in \varnothing \}$$

But no $U \in \varnothing$ even exist, so we must have $$\bigcup \varnothing = \varnothing$$

If it's the notation you're confused about, bear in mind that, for example, $$A \cup B = \bigcup \{ A, B \}$$ Hopefully some of the waffle I've given above clears up your confusion.

Clive Newstead
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  • That kind of answered my question, but would it be wrong to include $\emptyset \in B$? – Lemon Sep 13 '14 at 17:35
  • It would be wrong to conclude $\varnothing \in B$. $B$ has four elements, and none of them are $\varnothing$. (It is true that $\varnothing \subseteq B$ though.) – Clive Newstead Sep 13 '14 at 17:36
  • No, "include", not "conclude". So is it wrong to put the empty set inside $B$ is what I am asking. – Lemon Sep 13 '14 at 17:40
  • @sidht: Well what's $B$ meant to be? Are you asking if it becomes a basis when you put $\varnothing \in B$? – Clive Newstead Sep 13 '14 at 19:11
  • No, I am asking if it is okay to write $$B = { \emptyset, {a}, {c},{a,b},{ b,c} }.$$ – Lemon Sep 13 '14 at 22:08
  • @sidht: In that case my first comment did answer your question. It's false to say that, because $B$ is fixed, you're given what it is, it has four element and none of its elements are the empty set. – Clive Newstead Sep 14 '14 at 14:32
  • I know it's given. BUt I was asking if $B$ was given like what I wrote, is it still okay? – Lemon Sep 14 '14 at 18:28