1.Let $X$ be a set.
Let $\mathscr{A}$ be a family of subsets of $X$.
Here, what does 'a family' means precisely?
Does this mean $\{f(\alpha)\}_{\alpha\in A}$ for some $A\subset P(X)$ and $f:A\rightarrow P(X)$?
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A "family" of sets is just a set of sets. People use the word "family" or "collection" to avoid becoming confused or confusing others. So yes, $\mathcal{A}$ is a subset of the power set of $X$. – Stefan Smith Oct 30 '13 at 00:57
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Basically it means $\mathscr{A} \subset {\cal P}(X)$. – copper.hat Oct 30 '13 at 00:58
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@Stefan How do i interpret 'family' in first-order logic? – Jj- Oct 30 '13 at 00:58
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2There is no logic in my family. – copper.hat Oct 30 '13 at 00:59
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@copper.hat Thank you! However, here is an awkward example. As you know, $T$ is a topology on $X$ iff 1. $\emptyset \in T$ and $X\in T$ 2. Intersection of any finite family of sets in $T$ is an element of $T$ 3. Union of any family of sets in $T$ is an element of $T$. – Jj- Oct 30 '13 at 01:02
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From you definition, 3&2 imply 1, if we take $\bigcap \emptyset = X$.why do people generally require the first axiom? – Jj- Oct 30 '13 at 01:04
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1Because one doesn't want the understanding of a topology to depend on silly stuff about empty intersections. – André Nicolas Oct 30 '13 at 01:12
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1You're quite right that clause 1 in the definition of "topology" is redundant. It's included for the benefit of people who don't like to think about the empty family or who may be confused about what its union and (especially) its intersection are. – Andreas Blass Oct 30 '13 at 01:13
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They are redundant. Look at http://math.stackexchange.com/a/370201/27978. – copper.hat Oct 30 '13 at 01:21
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(1) There's nothing wrong with redundancy. Matter of taste. (2) If you want to get rid of redundancy in the definition of a topology, the way to do it is to change "any finite family of sets" to "any two sets". (3) Condition 2 in the definition of a topology contains a lot of redundancy all by itself. It says that, for any $n\in\omega$, the intersection of $n$ open sets is open. You really only need to assume this for $n=2$ (and $n=0$ if you want to be silly and discard condition 1). – bof Oct 30 '13 at 01:36
