Take $R$ to be the relation defined on $P(\{1, . . . , 100\})$ by $A \sim B$ if and only if $|A \cap B|$ is even.
Firstly, am I right to think that for example, $|\{0\}\cap \{1\}| = |\{1\}| = 1$.
And I'm not too sure how to prove the properties, this is my first time on relations on power sets. This is what I have:
$1.$ Not anti-symmetric: because $\{1\}\sim \{2\}$ and $\{2\}\sim \{1\}$ but $\{1\} \ne \{2\}$
$2.$ Transitive: Assumption is $a\sim b$, $b\sim c$. There are two cases where these are true. First is when $a,b,c$ are elements with an even cardinality. So if $a$ and $c$ are even, then $a\sim c$ since the sum of two even numbers is even. The other case is where $a,b,c$ are odd. The sum of two odd numbers is even.
$3.$ Reflexive: because if both elements on either side of the relation are equal, then it's $2$ times an element with an even or odd cardinality, which is equal to an even cardinality.
$4.$ I think it's symmetric but not sure how to prove $a\sim b$ implies $b\sim a$.
Sorry for the lengthy post, I hope this is fine with the rules. Appreciate any help on this topic!
No.
Firstly, it is improper to write "and" in this context. Use an intersection sign. $|\{0\}\cap \{1\}|$.
Second, $\{0\}$ is not an element of the powerset of $\{1,2,3,\dots,100\}$ since $0$ is not an element of $\{1,2,3,\dots,100\}$... so there is little point to have brought it up in an example.
Third, $\{0\}\cap \{1\}$ is equal to the emptyset $\emptyset$ since $\{0\}$ and $\{1\}$ share no elements in common. This of course implies that $|\{0\}\cap \{1\}|=0$
This is correct
While it is true that "even plus even = even" this has nothing to do with the setting here.
Consider the following: $\{1,2,3\}R\{2,3,4\}$ since they each share two elements with each other (an even number) and $\{2,3,4\}R\{3,4,5\}$ for the same reason (that is to say $|\{1,2,3\}\cap \{2,3,4\}| = |\{2,3\}|=2$ and so on...) however $\{1,2,3\}$ is not related to $\{3,4,5\}$ since these share only one element in common (and one is not even).
Even more directly, $\{1\}$ is related to $\emptyset$ and $\emptyset$ is related to $\{1\}$ however $\{1\}$ is not related to $\{1\}$
This is a very alarming statement, perhaps even moreso than your first. No, this is very incorrect and makes me believe you don't know what basic set operations like intersection mean. Perhaps you were confusing intersection $\cap$ with... disjoint union perhaps?
Take the example of $\{1\}$ and asking if it is related to itself. Here we have $\{1\}\cap \{1\}$. We ask ourselves... "what, if any, elements are in both of these sets?" The answer is that there is one element... the element $1$. That is, $\{1\}\cap \{1\} = \{1\}$. The number of elements in the intersection? $1$. We count it only the one time... not twice.
Similarly, $\{2,3,4,5\}\cap \{2,3,4,5\} = \{2,3,4,5\}$ has four elements in it. Here, we do have that $\{2,3,4,5\}$ is related to itself, but since we have examples of sets who are not related to themselves the relation as a whole is not considered to be reflexive.
It is well known that disjunction and conjunction is commutative. That is to say... "This and that" is the same as "That and this."
We know then that $A\cap B = \{x~:~x\in A~\text{and }x\in B\}$ is equal to $\{x~:~x\in B~\text{and }x\in A\}$ which we know to be equal to $B\cap A$. In other words, $A\cap B=B\cap A$ and so intersection is commutative as well.
If we know that $|A\cap B|$ is even and knowing that $|B\cap A|$ is equal to the same thing, it follows that $|B\cap A|$ is even as well and so $a\sim b$ does imply that $b\sim a$