Results 1 to 3 of 3

Math Help - Translating to set notation with existence.

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    13

    Translating to set notation with existence.

    Hey all,
    I understand basic set notation, but I'm having trouble understanding these 2 examples which deal with existence and translating them to set notation. If someone could explain them thoroughly , I'd appreciate it.
    Thanks in advance.

    1.There are some B that are not H.
    Answer:  B \bigcap H' \neq \emptyset

    2. \exists x. (x \in H \implies x \in B) \wedge (x \notin E)
    Answer: (H' \bigcup  B) \bigcap E' \neq \emptyset
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,561
    Thanks
    785

    Re: Translating to set notation with existence.

    What exactly do you need to do: translate these statements to set notation or explain why the first line and the second line say the same thing?

    Quote Originally Posted by cynlix View Post
    1.There are some B that are not H.
    Answer:  B \bigcap H' \neq \emptyset
    Let B be the set of male cats and H be the set of white cats. Then the complement of H, i.e., H', is the set of non-white cats, so B\cap H' is the set of male non-white cats. Saying that this set is nonempty is the same as saying that there exists a male cat that is non-white.

    In general, there is a direct correspondence between set operations on (the extensions of) properties and logical connectives. Let A be the set of objects having a property P and B be the set of objects having a property Q. In set notation, A=\{x\mid P(x)\} and B=\{x\mid Q(x)\}. The set A is called the extension of property P and similarly for B and Q. Then A\cap B=\{x\mid P(x)\land Q(x)\}, A\cup B=\{x\mid P(x)\lor Q(x)\} and A'=\{x\mid\neg P(x)\} where \land, \lor and \neg mean "and," "or," and "not," respectively. So, the laws like De Morgan's carry over from logic to set theory and back: (A\cap B)'=A'\cup B'=\{x\mid\neg(P(x)\land Q(x))\}=\{x\mid\neg P(x)\lor \neg Q(x)\}.

    Quote Originally Posted by cynlix View Post
    2. \exists x. (x \in H \implies x \in B) \wedge (x \notin E)
    Answer: (H' \bigcup  B) \bigcap E' \neq \emptyset
    It should be clear that H is the extension of the property x\in H, i.e., H=\{x\mid x\in H\}. Using the fact that R\implies S=\neg R\lor S and the correspondence described above, \{x\mid (x\in H\implies x\in B)\land x\notin E\}=\{x\mid (x\notin H\lor x\in B)\land x\notin E\}=(H'\cup B)\cap E'.

    Note that A\cap B (\cap in LaTeX) denotes the intersection of A and B, i.e., the set of common elements. In contrast, \bigcap A denotes the intersection of the elements of A when A is a family of sets. I.e., if A=\{a_1,\dots,a_n\}, then \bigcap A=a_1\cap\dots a_n. A similar thing holds for \cup (\cup in LaTeX) and \bigcup.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2010
    Posts
    1

    Re: Translating to set notation with existence.

    Alrighty, let me see if I can help you make some sense out of these.

    For the first one, you can think of as \exists x: x\in B \wedge  x\notin H

    Now think about taking the intersection of B and H compliment, or B \bigcap H' . Will it be empty? The answer is no. We defined the problem by saying that there was something in both! So we can write A \bigcap H'  \neq \emptyset

    If you understand that, then the second one can be done in a similar manner.

    Lets ask ourselves what x would satisfy that conditional. We know that a conditional is satisfied when either the antecedent ( p in  p \implies q) is false, or (inclusive!) the consequent (the q) is true. So we want x \in H to be false, which is equivalent to x \in H' being true. The consequent is that  x \in B . So we can really think about the conditional as being the same as H' \bigcup B Now consider the second part of the intial statement. It tells us that there is an x which satisifies the first conditon AND is also a member of E'. so if we take the original statement and intersect it with E, we get (H' \bigcup B ) \bigcap E' \neq \emptyset. Why? Because we said that there is an x that satisfies those conditions, so that set cannot be empty.

    Does that make sense, or am I just rambling?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Translating words into algebraic expressions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 12th 2012, 08:35 AM
  2. Translating English to FO Logic
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: August 3rd 2010, 07:32 AM
  3. Predicate calculus translating sentences
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 5th 2009, 03:54 PM
  4. Translating a polygon
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: January 5th 2009, 04:56 PM
  5. Need help translating idea in math formula
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: March 12th 2007, 12:34 PM

Search Tags


/mathhelpforum @mathhelpforum