0
$\begingroup$

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 $X$?

$\endgroup$
  • $\begingroup$ 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." $\endgroup$ – Total Anime Immersion 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. $\endgroup$ – Total Anime Immersion 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Total Anime Immersion 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? $\endgroup$ – Total Anime Immersion 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? $\endgroup$ – Total Anime Immersion Jul 11 '14 at 19:34
2
$\begingroup$

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 $\bigcup \emptyset = \emptyset$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Total Anime Immersion 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? $\endgroup$ – Total Anime Immersion Jul 11 '14 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.