Proof of sets. Need an example

Question

My question is to show that $X-(Y \cup Z)$ is a subset of $(X-Y) \cup (X-Z)$. I already did the proof for that and understand that but the second part is to give an example to show that in general, the sets are not equal. I don't know what kind of example that would be and need some help with this please!

Answer 1

Hint:

Where

Blue: $X-(Y\cup Z)$

Yellow: $(X-Y)\cup (X-Z)$

Answer 2

Find an element, $e$, of $Z$ and in $X$ that is not in $Y$. $e$ is in $X-Y$ so $e$ is in $(X-Y)\cup(X-Z)$.

But $e$ is $Z$, so $e$ is in $Y\cup Z$ so $e$ is not in $X-(Y\cup Z)$. So a proper subset.

So as long as $X$ intersects $Y$ or $Z$ and that intersection has is not a subset of $Z$ or $Y$, it will be a proper subset.

Try $X = \{0, 1\}, Y = \{1, 2\}, Z = \{2, 3\}$ [$1$ will be in the larger set but not the subset]

Answer 3

Try the case where $X$ is any non-empty set and then let $Y=X$ and $Z=\varnothing$. In this case $X-(Y \cup Z) = \varnothing$ while $(X-Y) \cup (X-Z) = X$.

Answer 4

Sorry for double answering but:

$(X-Y) \cup (X-Z) $ is all X that are not both Y and Z. $X-(Y \cup Z) $ is all X that are neither Y nor Z.

So samples work with just about any sets you want.

X = all toys

Y = all red things

Z = all plastic things.

$(X-Y) \cup (X-Z) $ = all toys that aren't red plastic. $X-(Y \cup Z) $= all toys that are neither red nor plastic.

Or X = all positive integers

Y= all even numbers

Z = perfect squares

$(X-Y) \cup (X-Z) $= all odd numbers U all numbers that aren't squares = all numbers that aren't even squares = {1,2,3,5,6,...}

$X-(Y \cup Z) $ all numbers that neither even nor perfect squares = all odd non-squares = {3,5, 7, 11,..}

Answer 5

For an extreme example, let $Y$ and $Z$ be any nonempty sets and let $X=Y\cup Z$. On the one hand we have $X-X=\emptyset$ but $(X-Y)\cup(X-Z)=Z\cup Y=X$.