why $\bigcap\mathscr A= A ?$

Question

Taken from General Topology by stephen willard

My confusion given in red box

My attempt : here its given $\bigcap\mathscr A= A$

I think it should be $\bigcap\mathscr A = \emptyset$

Note : Im not able to write that letter in mathsjax

See here https://math.stackexchange.com/questions/370188/empty-intersection-and-empty-union — Sumanta, Sep 12 '20 at 07:15
@jasmine Best thing to do is to forget about this (beyond) annoying convention and simply renounce the idea of handling empty intersections, unless you were working in some version of the Gödel-Bernays axiomatic system, in which case $\bigcap \varnothing=\mathbf{\mathscr{U}}$ (the whole universe). — ΑΘΩ, Sep 12 '20 at 07:49
@ΑΘΩ I disagree. The convention is quite handy. Especially in topology. As soon as you are master over it you start enjoying that you did not put is aside and gain understanding why it is accepted as convention. — drhab, Sep 12 '20 at 08:47
@jasmine Be aware of this: the "smaller" $\mathscr A$ the "larger" $\bigcup\mathscr A$. Some candidate $x$ will be an element of $\mathscr A$ if it satisfies the following conditions: $x\in A$ for every $A\in\mathscr A$. The smaller the number of conditions the larger the number of candidates $x$ that pass this test. That is why thinking that $\bigcap\mathscr A=\varnothing$ is actually counterintuitive. In that case there is no condition at all, so every candidate $x$ passes the test. — drhab, Sep 12 '20 at 08:59
@drhab Sir, you are entitled to your opinion, of course. As far as I am concerned, it is not a question of mastering anything -- it is really not that complicated a convention at all to require any mastery of any sorts -- as it is a question of being revolted -- on the grounds of a very firm espousal of the principles of syntactical and semantic rigour -- at the notion that one and the same entity constantly denoted by $\bigcap \varnothing$ could mean this ambient set here in this context or that thing there in that context (to be cont.) — ΑΘΩ, Sep 12 '20 at 13:32
@drhab (cont.) Up to the point where -- if one were to take this syntax to its logical consequences -- one would derive the conclusion that any two objects $x$ and $y$ are equal, since any set can serve as an ambient set within which to consider the empty intersection of (none of) its subsets.....From such a position of rigour, I might agree with a version of the notation where a syntactic amendment such as $\bigcap_{\mathscr{A}}\varnothing$ were adopted, to indicate the fact that one is working in the ambience of $\mathscr{A}$. (to be cont.) — ΑΘΩ, Sep 12 '20 at 13:35
@drhab (cont.) However, as things stand, I personally treasure the clarity and correctness of syntax rather than any allegedly advantageous conventions which are ultimately syntactically incorrect and illicit. — ΑΘΩ, Sep 12 '20 at 15:12

score 3 · Accepted Answer · 2020-09-12T07:30:28.977

The assertion $\bigcap \emptyset=\emptyset$ is unwarranted because it is quite clear that the condition $[\forall y\in\emptyset,\ x\in y]$ is satisfied for all objects $x$. Now, in these terms, the intersection of (all the elements of) the empty set should be the class of all sets. However, there is a valid argument for assuming implicitly that, when an ambient set $U$ is obvious from context, the notation $\bigcap \mathscr A$, for some $\mathscr A\subseteq \mathcal P(U)$, stands for $\{x\in U\,:\, \forall y\in\mathscr A, x\in y\}$. In that sense, $\bigcap \emptyset=U$.

score 2 · Answer 2 · answered Sep 12 '20 at 07:23

For subsets of $A$ and $\mathcal{A}$ a collection of subsets of $A$ we define

$$\bigcap \mathcal{A}=\{x \in A\mid \forall C \in \mathcal{A}: x \in C\}$$

And when $\mathcal{A}=\emptyset$ the $\forall$ statement is always true (void truth) so the statement after the $\mid$ holds for any $x \in A$. So $\bigcap \emptyset = A$.

Note that this is an ugly "trick": if we did not specify the "universe" to be $A$ (but $B \supseteq A$ e.g.) this same intersection would have been $B$ instead. Or in naive set theory e would get "the set of all sets" which cannot exist (or co-exist with comprehension, rather) by Russell's paradox. It's a handy convention in topology, as it allows any collection of subsets to be a subbase for a topology (later in the book).

For now, I'd accept this reasoning and use it as a handy convention.

score 2 · Answer 3 · answered Sep 12 '20 at 09:31

The following conversation concerns any $x\in A$.

"Sorry, sir. I have some $x\in A$ here and he wants to know whether he is an element of $\bigcap\mathscr A$."

"For that it must be checked whether there is some $U\in\mathscr A$ such that $x\notin U$. Such $U$ would prevent it."

"But there is no $U$ that satisfies $U\in\mathscr A$."

"Fine, then there is no veto, hence he is indeed a member of $\bigcap\mathscr A$."

why $\bigcap\mathscr A= A ?$

3 Answers3