-1
$\begingroup$

Problem:

Sets and Symmetry. Prove

I know a) is True because if A = null set then we can take B = {271}

b) the set difference of A is a null set and therefore it is False

c) is also true if you subtract A from both sides (aka symmetric difference of A from both sides). Then you are left with B = C.

I am however having trouble how to mathematically prove these 3 for a paper to receive full grades. Can someone please show the method.

Thanks

  • 1
    Your reasoning for $a)$ is incorrect. That only proves one case of $A$ (the case where $A = \emptyset$)2017-02-16

1 Answers 1

0

For (a) you need to show it holds for all $A$, so you cannot provide a single example. You can break it into two cases. First, let $271\in{}A$ and you can select a set $B$ such that $271\notin{}B$. Then the next case is that $271\notin{}A$, then you can select $B$ such that $271\in{}B$. This shows that it is true for all $A$.

For (b), you claim an exists statement is false, therefore you must show the opposite is true for all $B$. You just need to show that for all $B$, you can find an $A$ such that $271\notin{}A\Delta{}B$. This again can be done by splitting into the two cases $271\in{}B$ and $271\notin{}B$ and proceeding in a similar manner as above by finding appropriate $A$.

For (c) you can do the following argument. Take $x\in{}A\Delta{}B$, then either $x\in{}A-B$ or $x\in{}B-A$. If $x\in{}B-A$, then we also know that $x\in{}A\Delta{}C$, so since $x\notin{}A$, then it must be the case that $x\in{}C$. If on the other hand we have that $x\in{}A-B$, then it must be the case that $x\notin{}C$ from same line of reasoning.