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I am attempting to work on some proofs for my math assignment, but I'll be honest in that I am really struggling to understand them. I read through the power point given by my teacher; however, even after asking for help I'm not really understanding why the proofs work the way they do. I know this might be a lot to ask, but I would love if someone could maybe help me work out these problems or give me some guidance on how to approach them.

Let $A,B$ and $C$ be sets. Show that

a) $(A\cup B)\subseteq (A\cup B \cup C)$
b) $(A\cap B \cap C)\subseteq (A\cap B)$
c) $(A- B)-C\subseteq A-C$
d) $(A- C)\cap (C-B)=\emptyset$
e) $(B-A)\cup (C-A)=(B\cup C)-A$

I am on the first one still I currently have: $x \in A \cup B$, so $x \in A$ or $x \in B$.

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  • $\begingroup$ you can make a start by considering an arbitrary element of left set and show that it is or isn't contained in the right set $\endgroup$ – danimal May 21 '15 at 16:14
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    $\begingroup$ To show two sets are equal: Prove $S \subseteq T$ and $T \subseteq S$. To show something is a subset: Assume $x \in S$ and then prove that $x \in T$ also. This is the standard way of proving subset relations. $\endgroup$ – user222031 May 21 '15 at 16:15
  • $\begingroup$ Venn Diagrams can be helpful guides for figuring out how the proofs will work. With a little formalization, Venn Diagrams can essentially be proof! $\endgroup$ – Bob Krueger May 21 '15 at 17:07
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(a): Suppose $x\in A\cup B$. Then $x\in A$ or $x\in B$. If $x\in A$, then $x\in A\cup B\cup C$. If, however, $x\in B$, then $x\in A\cup B\cup C$. Either way, $x\in A\cup B\cup C$ when $x\in A\cup B$. Thus, $A\cup B\subseteq A\cup B\cup C$.

(b): Suppose $x\in A\cap B\cap C$. Then $x\in A$ and $x\in B$ and $x\in C$. Hence, we have that $x\in A$ and $x\in B$; that is, $x\in A\cap B$. Thus, $A\cap B\cap C\subseteq A\cap B$.

(c): Suppose $x\in (A-B)-C$. Then $x\in A-B$ and $x\not\in C$. That is, $x\in A$ and $x\not\in B$ and $x\not\in C$. Hence, $x\in A$ and $x\not\in B$; that is $x\in A-C$. Thus, $(A-B)-C\subseteq A-C$.

(d): Using some set algebra (you can use element-chasing proofs as above, but I would advise against this when simple set algebra can take care of everything), where $S^C$ denotes the complement of the set $S$, we have the following: \begin{align} (A-C)\cap(C-B) &= (A\cap C^C)\cap(C\cap B^C)\tag{definition}\\[0.5em] &= (C^C\cap C)\cap(A\cap B^C)\tag{associativity}\\[0.5em] &= \varnothing\cap(A\cap B^C)\tag{$S-S=\varnothing$}\\[0.5em] &= \varnothing. \end{align} (e): \begin{align} (B-A)\cup(C-A) &= (B\cap A^C)\cup(C\cap A^C)\tag{definition}\\[0.5em] &= (B\cup C)\cap A^C\tag{distributivity}\\[0.5em] &= (B\cup C)-A\tag{definition} \end{align}

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  • $\begingroup$ Incredibly helpful and clear. I am actually able to do the others now. Unsure as to why this was confusing to me, but thank you for your time and help. You really improved my day! $\endgroup$ – Moxy May 21 '15 at 16:54
  • $\begingroup$ @Moxy Glad it helped! In the future, I would suggest adding more to your question in terms of your own thoughts/work. Also, this question was really five questions in one. You should really split them up separately when they do not depend on each other, as they do not here. Also, you really should see this tutorial on typesetting. It will make your posts easier to read and better received in the future. Cheerio. $\endgroup$ – Daniel W. Farlow May 21 '15 at 16:56
  • $\begingroup$ Got it! Thank you for the tips. I will remember them for future posts! $\endgroup$ – Moxy May 21 '15 at 16:58

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