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I don't really know what to do with unions, can someone help me?

I assume $B\subseteq C$, (I want to show that $A\cup B \subseteq A\cup C$) suppose $A\cup B \subseteq A\cup C$ and let $x\in A\cup C$ so that

$x\in A$ or $x\in C$...

I don't really know what to do up to this point, can someone offer me guidance?

John Griffin
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Ryan Pang
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  • Ah... well... with the edit now, it is a true claim unlike before. Suppose $x\in A\cup B$, then by definition $x\in A$ or $x\in B$. In the first case where $x\in A$ is it true that $x\in A\cup C$? In the second case where $x\in B$ is it true that $x\in A\cup C$? (here the fact that $B\subseteq C$ comes into play) – JMoravitz Aug 19 '17 at 00:12
  • Thanks for responding to my question, I mixed up what I was assuming with what I wanted to prove. Again, thanks for the help. – Ryan Pang Aug 19 '17 at 00:17

3 Answers3

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You just want to prove that every element in $A\cup B$ must be in $A\cup C$. To this end, suppose $x\in A\cup B$. By the definition of union, this means that $x\in A$ or $x\in B$. Let's consider these cases separately:

If $x\in A$, then $x\in A\cup C$.

If $x\in B$, then from $B\subseteq C$ we obtain $x\in C$ and consequently $x\in A\cup C$.

Therefore every element in $A\cup B$ is in $A\cup C$, so we have $A\cup B\subseteq A\cup C$.

John Griffin
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  • Thanks for the reply, I have a poor understanding of unions but your response definitely helps! – Ryan Pang Aug 19 '17 at 00:15
  • @RyanPang The union of two sets $A$ and $B$ is just the set of all elements that are in either $A$ or $B$ ("or" here includes the possibility that it's in both). In symbols, it is: $$ A \cup B = {x \mid x\in A\ \text{or}\ x\in B}. $$ The key fact is that $x\in A\cup B$ if and only if $x\in A$ or $x\in B$. So knowing that something is in a union means we can assume it's in one of the sets, and if we want to prove that something is in a union, we just need to prove that it's in one of the sets. – John Griffin Aug 19 '17 at 00:16
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In his comment, OP Ryan says

Thanks for the reply, I have a poor understanding of unions but your response definitely helps!

To promote understanding and intuition, here is a more wordy/elaborate description of what is happening. We look at

$\tag 1 B\subseteq C \Rightarrow A\cup B \subseteq A\cup C$

as a way of modifying both $B$ and $C$. In a rigorous treatment you can't say things like 'a new $B$'.

We start with

$\tag 2 B \subseteq C$

and want to find a relationship between $A\cup B$ and $A\cup C$. But the union operation is how we 'add stuff into a set'. So, anything that gets added to both the left hand side $B$ and right hand side $C$ of (2) would not change the subset relationship relationship, since it would be 'added' to both sides.

In what follows we breakdown how each element of $A$ gets added to $B$ and $C$.

If $A$ is a subset of $B$, nothing changes using the union operation,

$\tag 3 A \subseteq B \,\land\, B\subseteq C \;\Rightarrow A\cup B = B \,\land\, A\cup C = C$

To show that (3) is true you need to show that $A$ is also a subset of $C$.

So how do you add 'new stuff'? One way is if $B$ is a proper subset of $C$, we can take an element $c$ that is in $C$ but not in $B$, and then

$\tag 4 \{c\} \cup B \subseteq \{c\} \cup C = C$

So when we add this kind of new stuff, nothing happens to $C$ but the 'new' $B$ 'looks more like' $C$.

The last thing to consider is adding new stuff that does not appear in $C$ (and then it also can't be in $B$). So if $k$ is a completely new object, we can throw that in also, and

$\tag 5 \{k\} \cup B \subseteq \{k\} \cup C$

In this case we changed both $B$ and $C$ by adding in a new element.

In summary, if you look at (1) in a dynamic way, then as you "pour" $A$ in,

(a) $B$ and $C$ don't change
(b) $B$ grows larger but remains a subset of $C$
(c) Both $B$ and $C$ grow larger in the same way
(d) A mixture of (a), (b), and (c)

CopyPasteIt
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Here is a more algebraic kind of proof:

In general you have that $X \subseteq Y$ iff $X \cap Y = X$ (take a minute to see why this is true)

Applied to your case:

We are given $B \subseteq C$, and so $B \cap C =B$

Therefore, $(A \cup B) \cap (A \cup C) = A \cup (B \cap C) = A \cup B$

So, since $(A \cup B) \cap (A \cup C) = A \cup B$, we have $A \cup B \subseteq A \cup C$

Bram28
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