# Munkres Topology Ch 1 ex#7) part (c) — basic set theory Q

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## Homework Statement:

This is Munkres Topology ch 1, exercise #7: Write the given set in terms of the sets ##A, B, C## and the symbols ##\cup , \cap , \text{ and } -##.

$$F=\left\{ x| x\in A \wedge \left( x\in B\Rightarrow x\in C\right) \right\}$$

## Relevant Equations:

DeMorgan’s laws perhaps? Idk.
Obviously the parenthetical part of the definition of ##F## means ##B\subset C## but we are not allowed to use ##\subset##. I do not know how to express implication with only union, intersection, and set minus without the side relation ##B\cap C = B\Leftrightarrow B\subset C##. This is using the correct symbols but I think he wants a single relation. The “and” part is intersection of A with B but this doesn’t convey that B is a subset of C. This is going to be simple I bet.

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Math_QED
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Hint: Given two logical formulas ##p## and ##q##, we have

$$(p \implies q) \equiv (\neg p \lor q)$$

Check with a truth table if you don't know this.

Then you can use De Morgan's laws.

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Thanks, apparently during the 20 years since I took logic, I’ve forgotten a few things. Lol

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Ok, so I get $$F=\left\{ x | x \in A \wedge ( x \not\in B \lor x \in C ) \right\} = A\cap C$$ but I didn’t use DeMorgan’s Laws, I just drew a picture. I figure picture drawing with only work for “student problems” though, so I need to learn to do this symbolically. I suppose I’ll need to call the universe ##X## so I can deal with the ##x \not\in B## part, so that $$F = A\cap [ ( X - B ) \cup C ]$$ and I don’t see how to manipulate this into the form of DeMorgan’s Law which would require the union or intersection of two differences of sets within the square brackets.

Note: I’m just hoping my TeX processes correctly bc for some reason the preview isn’t displaying the symbols on my phone, just the code.

Math_QED
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$$F = A\cap [ ( X - B ) \cup C ]$$
Strictly speaking that's already correct. You can use De Morgan to get an equivalent expression but that's not really necessary.

Last edited:
benorin
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Ok, so I get $$F=\left\{ x | x \in A \wedge ( x \not\in B \lor x \in C ) \right\} = A\cap C$$ but I didn’t use DeMorgan’s Laws, I just drew a picture.
What if ##x## is in ##A## but not in ##B## and also not in ##C##?

benorin
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What if ##x## is in ##A## but not in ##B## and also not in ##C##?
Oh thanks, missed that region! It should be ##F=\left\{ x | x \in A \wedge ( x \not\in B \lor x \in C ) \right\} = (A\cap C)\cup (A-B)##. You guys are awesome! Always giving the right amount of help and not just solving the problem for me. I'm taking mental notes on how to teach from you guys, it's been years since I tutored or taught and working here is bringing things back for me.

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My pleasure! One other thing I will point out that might make your life a little easier when dealing with these types of questions:
##\wedge## and ##\cap## both point up.
##\vee## and ##\cup## both point down.

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mathwonk
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If it makes you feel any better, I am a professional mathematician, and I barely could do this, even after some time, and it does not bother me one whit, since doing this is totally unrelated to doing math. When I write a math paper I do not try to make it hard for people to understand what I am saying by using symbols when words would be clearer. Still if you enjoy this sort of game, that's great.

Math_QED
Math_QED
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If it makes you feel any better, I am a professional mathematician, and I barely could do this, even after some time, and it does not bother me one whit, since doing this is totally unrelated to doing math. When I write a math paper I do not try to make it hard for people to understand what I am saying by using symbols when words would be clearer. Still if you enjoy this sort of game, that's great.
With all the respect, but I don't buy that a professional mathematician can't do that. Surely you must be able to do set manipulation? I.e. De Morgan's laws etc.

mathwonk
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Forgive me, I am getting old and dull. This is indeed a puzzle I would have enjoyed as a young man.

George Jones
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Given two logical formulas ##p## and ##q##, we have

$$(p \implies q) \equiv (\neg p \lor q)$$
I find the equivalent statement
$$\neg \left( p \implies q \right) \equiv \left( p \land \neg q \right)$$
to be more "intuitive".

Math_QED
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I find the equivalent statement
$$\neg \left( p \implies q \right) \equiv \left( p \land \neg q \right)$$
to be more "intuitive".
I agree, it is not what the OP needed though, but it's how I 'remember' it as well. I simply took the negation of it.