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Proof

I have no idea what is being asked here. What is the a, and what does it mean that they are indexed by I? Also, why are they a and the c stacked?

banana
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    Indexed here simply means the subsets $V$ are being labelled by elements $\alpha$ of $I$. Also, the $c$ indicates the compliment of the set. – 高田航 Apr 18 '17 at 22:28

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Being indexed by $I$ means that there is a bijection between the elements of $I$ and the sets you are unioning or intersecting over. The $\alpha$ is just a dummy variable. For example, the family of sets $\{x \}$ where $x \in \mathbb{R}$ is a family of sets, and this family is indexed by the real numbers, because for every real number, there is the set $\{x\}$. The union of such sets is the whole line. Since each is disjoint from the others, the intersection of all these sets is empty.

The $c$ above means to take the complement. The result you are asked to prove is known as De Morgan's Rule.

A. Thomas Yerger
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  • So, I'm being asked to prove that the complement of the union of all the sets is equal to the intersection of the complement of V? – banana Apr 18 '17 at 22:40
  • @JerseyFonseca Correct. You can convince yourself this is true by a picture say with three sets, and then use this new notation to handle the arbitrary case. – A. Thomas Yerger Apr 18 '17 at 22:41
  • Thank you so much. And I'm very sorry, but my definition has a faulty definition of complement that I don't understand very well. (I'm struggling very much in this class). What exactly is the complement? I'm having a hard time fundamentally understanding – banana Apr 18 '17 at 22:46
  • Given a set $A$, $A^c={x: x\not\in A}$. – Alekos Robotis Apr 18 '17 at 22:49
  • Okay. Final question: How do I prove that all of these are disjointed from each other? – banana Apr 18 '17 at 22:51
  • @JerseyFonseca they may not be disjoint in general. Just draw two blobs in the plane that overlap. It is still true that the complement of the intersection is the union of the complements. You should do some informal examples of this type just with blobbies until you're comfortable with what you're trying to prove. – A. Thomas Yerger Apr 18 '17 at 22:53
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Of course, without more context it isn't obvious what $I$ is, but it is clearly some set. $\{V_\alpha\}_{\alpha \in I}$ simply denotes that each of the $V_\alpha$ is indexed by an element of $I$. If you like, we construct a bijective function from $I$ to our collection $\{V_\alpha\}$. Another example of this is the integer indexed collection of intervals: $$ \{J_n\}_{n\in \mathbf{Z}}$$ where $J_n=(n,n+1)$ for $n\in \mathbf{Z}$. In this case, $V_\alpha^c$ denotes the complement of the element of $\{V_\alpha\}_{\alpha \in I}$ indexed by $\alpha$.

Alekos Robotis
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