I was reading Introduction to Topology by George L. Cain and found myself struggling with this definition mentioned in the book.

Let $X$ be a set, and suppose $C$ is a collection of subsets of $X$. Then if $C$ = Empty Set, Union of $C$ is an Empty Set too and Intersection of $C$ is the set $X$.

Now my questions are:

  1. If $C$ is an Empty Set and also the collection of subsets of $X$ then isn't it true that $X$ is also essentially an empty set. For example, let $X = \{1,2\}$ then according to the definition $C = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$. So, in the same manner $C$ can be an Empty Set only when the collection of subsets of $X$ is an Empty Set or $X = \emptyset$.

  2. Union of $C$ is the union of all the elements present in $C$. So if $C$ is an Empty Set, then the only set present in the collection of sets $C$ is the value of an Empty Set. Here lies my question: We know how to find the Union of $2$ or more sets but with respect to what should I find the Union of $1$ set? Also, at the back of my mind I know its an Empty set because there aren't any other sets present in $C$ but how do I know for sure?

  3. Intersection of $C$ is the intersection of all the elements present in $C$. So if $C$ is an empty set, again I present the same question, with respect to what should I take the intersection? Also, generally if $C$ was not an empty set and would be something like $C = \{\emptyset, \{1\}\}$, the intersection would be equal to an Empty Set as the set $\{1\}$ has subsets $\{1\}, \emptyset$. So how exactly does the Intersection of $C$ when $C$ is an Empty Set return set $X$?

  • $\begingroup$ Are you sure the text reads "Let $X$ be a set and $C$ be the collection of its subsets"? $\endgroup$
    – Git Gud
    Jul 11 '14 at 16:30
  • $\begingroup$ The text reads so, "Let X be a set, and suppose C is a collection of subsets of X." $\endgroup$ Jul 11 '14 at 16:31
  • $\begingroup$ The use of the definite article 'the' as in the question makes a huge difference and is part of your confusion. $\endgroup$
    – Git Gud
    Jul 11 '14 at 16:34
  • $\begingroup$ I didn't understand. Could you please elaborate what you mean? Thank you. $\endgroup$ Jul 11 '14 at 16:36
  • $\begingroup$ I was in the process of typing an answer, but two users got ahead of me. I hope your confusion will now be clarified. $\endgroup$
    – Git Gud
    Jul 11 '14 at 16:52

Intuitively, when the collection $\mathscr C$ grows larger, its union grows larger and its intersection grows smaller. Going the other way, when the collection $\mathscr C$ grows smaller, its union grows smaller and its intersection grows larger. Taken to the extreme, when the collection $\mathscr C$ is as small as possible (empty), its union is as small as possible (empty) and its intersection is as large as possible (the whole space).

More technically, a point $x\in X$ is in $\bigcup \mathscr C$ if there is a member $C$ of the collection $\mathscr C$ with $x\in C$. When $\mathscr C$ is empty, this can't happen, so no point qualifies to be in $\bigcup \mathscr C$.

Similarly, a point $x\in X$ is in $\bigcap \mathscr C$ if $x\in C$ for every member $C$ of the collection $\mathscr C$. When $\mathscr C$ is empty, this is vacuously true (you can't demonstrate a member of $\mathscr C$ that fails to contain $x$).

So, the union of a family "starts out empty" and grows as you add sets to the family (more points qualify to belong), and the intersection of a family "starts out universal" and shrinks as you add sets to the family (fewer points qualify to belong).

Edit: I can't comment any more or the comments will be moved to chat; so I will "cheat" and comment here (sorry for the breach of protocol). No, it doesn't go beyond $X$ itself because we specified that $\mathscr C$ was a collection of "subsets of $X$" to begin with. I know that's a little vague, but $X$ is the universal set in this context.

  • $\begingroup$ Reading the other answer and then looking at yours I perfectly understand the Union problem I faced and even the powerset and a part of collection of sets confusion I had but I didn't quite understand why every x qualifies for member of C in the intersection part? Could you explain a little more? $\endgroup$ Jul 11 '14 at 18:24
  • $\begingroup$ The important thing to see is the phrase "if $x\in C$ for every member $C$ of the collection $\mathscr C$". It is true because there's no member $C$ to start with. It's like me saying "Every pig dancing on top of my head is wearing blue socks". That is a true statement. It's not saying that there is a pig dancing on my head, it's saying that if there is a pig dancing on my head, then it is wearing blue socks. Implications of the form "false $\implies$ true" are always true. See? $\endgroup$
    – MPW
    Jul 11 '14 at 19:11
  • $\begingroup$ Ok, I get your point, although not completely. I understand that there are no members in the collection of the sets, now my question is, if there are no members, then Empty Set is also not a member of the collection. If that is so, then with what are we checking the factor of x belonging to the collection of sets? And if in case empty set is a part of the collection of the set(which it shouldn't be according to me, after trying to understand your explanation thoroughly) then any valid point x won't be present in Empty set, hence the answer should be an empty set right? $\endgroup$ Jul 11 '14 at 19:23
  • $\begingroup$ The collection $\mathscr C$ is itself empty. This means $C\in\mathscr C$ is false. Therefore the statement "If $C\in\mathscr C$, then $x\in C$" is true, because the "if" part is false. $\endgroup$
    – MPW
    Jul 11 '14 at 19:28
  • $\begingroup$ Okay, I had one last question, I hope I'm not disturbing you, and thanks for your time firstly. The definition says x belongs to X such that x belongs to C for every C belonging to the collection. Now my point is, if every x qualifies for this definition, then it should extend beyond the bounds of set X and should be the universal set U consisting of all values x which are both in set X and also those present in X'. Did I get the definition wrong or is it another misconception of a kind? $\endgroup$ Jul 11 '14 at 19:34

First, there is a big difference between a and the:

  • By the collection of subsets of $X$, we denote the so called power set of $X$, i. e. the collection of all subsets of $X$, for example if $X = \{1,2\}$, then the collection of subsets of $X$ is $\bigl\{\emptyset, \{1\}, \{2\},\{1,2\}\bigr\}$
  • By a collection $C$ of subsets of $X$ we denote an arbitrary collection of subsets, which may contain all subsets, but need not to. So $\{\{1\}\}$ is a collection of subsets of $\{1,2\}$, the whole powerset is, but $\emptyset$ is also, it is the collection of no subsets.

Regarding the union and the intersection. For a collection $C$ of subsets of $X$, one has ($X$ is fixed) $$ \bigcap C = \{x \in X \mid \forall A \in C: x \in A\} $$ and $$ \bigcup C = \{x \in X \mid \exists A \in C: x \in A \} $$ If now $C = \emptyset$ (the collection of no subsets), then $$ \bigcap \emptyset = \{x\in X \mid \forall A \in \emptyset: x \in A\} $$ As there is no element of $\emptyset$, every assertion is true on all its elements, so $\bigcap \emptyset = X$. $$ \bigcup \emptyset = \{x\in X \mid \exists A \in \emptyset : x \in A\}$$ As the emptyset has no elements, for no $x \in A$ there exsists an $A \in \emptyset$, hence $\bigcup \emptyset = \emptyset$.

  • $\begingroup$ About your assertion being true for the intersection: A should belong to Empty Set for every x which belongs to A, how is that true? As A is a non empty set. $\endgroup$ Jul 11 '14 at 17:15
  • $\begingroup$ I mean how can every A belong to Empty Set? In case of Intersection? And how can no A belong to Empty Set in Union? $\endgroup$ Jul 11 '14 at 17:22

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