When nothing is everything in Set theory

In an earlier post, I discussed the basic and most important aspects of Set theory, Functions and Real Number System. In the same, there was a significant discussion about the union and intersection of sets.

Restating the facts again, given a collection $ \mathcal{A}$ of sets, the union of the elements of $ \mathcal{A}$ is defined by

$ \displaystyle{\bigcup_{A \in \mathcal{A}}} A := {x : x \in A \textrm{ for at least one } A \in \mathcal{A} }$ .

The intersection of the elements of $ \mathcal{A}$ is defined by

$ \displaystyle{\bigcap_{A \in \mathcal{A}}} A := {x : x \in A \textrm{ for every } A \in \mathcal{A} }$ .

But what if the collection is empty? i.e., what if the collection contains no sets?

From the first day at Topology class in Gorakhpur University , it was the very first confusion which troubled many of classmates. Topology’s Professor was elaborating the finite intersection of sets in reference to topological space — and everyone in the classroom was enjoying it. To end of the lecture, as he wrote,

the intersection of members of empty collection equals to the universal set.

Which mathematically, Professor wrote $ \bigcap \emptyset = X$ where $ X$ is universal set. The quote was okay and significant but the mathematical notion of the same was confusing, as he represented an empty collection by “empty-set notation $ \emptyset$ ” . Professor easily asserted this to be an agreement among some mathematicians and even though the result is useful in some problems, he skipped a proper proof of it. [That’s really bad, Sir!]  At first instant, this problem lead me to a single result: $ \bigcap {\emptyset} = \emptyset$ , if $ \emptyset$ was usual null-set. But in the case, where $ \emptyset$ was an empty collection – I couldn’t reach to any result.

There came Munkres’s Topology book to the rescue. It’s second edition has a little note about the same on Chapter 1 page 12 .  He writes:

There is no problem with these definitions if one of the elements of $ \mathcal{A}$ happens to be the empty set. [In each such case of arbitrary union  and intersection the results is $ \emptyset$ .] But it is tricky to decide what (if anything) these definitions mean if we allow $ \mathcal{A}$ to be the empty collection. Applying the definitions literally, we see that  no element $ x$ satisfies the defining property for the union of the elements of $ \mathcal{A}$ . So it is reasonable to say that

$ \displaystyle{\bigcup_{A \in \mathcal{A}}} A := \emptyset$ if $ \mathcal{A}$ is empty.

On the other hand , every $ x$ satisfies (vacuously) the defining property for the intersection of the elements of $ \mathcal{A}$ . The question is, every $ x$ in what set? If one has a given large set $ X$ that is specified at the outset of the discussion to be one’s “universe of discourse”, and one considers only subsets of X throughout, it is reasonable to let

$ \displaystyle{\bigcap_{A \in \mathcal{A}}} A := X$ when $ \mathcal{A}$ is empty.

Not all Mathematicians follow this convention, however. …

 

So, according to Munkres (with some very good arguments), the intersection of elements of an empty collection of subsets of a set (say) $ E \subseteq X$ equals to the universal set $ X$ . In other words, nothing is everything. There must be any theoretical or logical proof of it! There must be.

Gaurav Tiwari

A designer by profession, a mathematician by education but a Blogger by hobby. With an experience of over seven years with WordPress, PHP and CSS3, Gaurav is capable of doing almost anything related to these. Beyond that, He is a mathematics graduate & a civil service aspirant.

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