# Prove that if $A\subseteq B$ then $A\setminus C\subseteq B\setminus C$

How can I prove that if $$A \subseteq B$$ then $$A\setminus C \subseteq B \setminus C$$?

I'm a bit lost. I have an idea that follows like so:

Let $x \in A, x \notin C$, then $\{x\} \in P(A\setminus C); \{x\} \in P(B\setminus C)$, while saying that nullsets are subsets of nullsets if $x \in C$.

However, I believe that there is a more straightforward solution. How should I prove the statement above?

This may make it somewhat easier for you to think about: you are given that $A\subseteq B$, from which you are to show that $A-C\subseteq B-C$ follows (that is, $A\subseteq B$ implies that $A-C\subseteq B-C$). Well, what does it mean for $A\subseteq B$ to be true? It means that "if $x\in A$, then $x\in B$." Alright now let's use this given information to show that the next part follows.

Suppose $x\in A-C$. Then this means that $x\in A$ and $x\not\in C$. Oh but wait--given that $A\subseteq B$, this means that if $x\in A$ then $x\in B$; thus, if $x\in A$ and $x\not\in C$, then this must mean that $x\in B$ also. Since you have that $x\in A$ and $x\not\in C$, then you must have $x\in B$ and $x\not\in C$ as well. Thus, you can see that if $A\subseteq B$ is given then $A-C\subseteq B-C$ follows.

Does that make more sense? It's a very dirty explanation, but it may be more intuitive for you to look at it that way.

• Every answer were really good, and I had a hard time picking which one to accept, but I find your explanation really helpful in getting me on the right track! – Secret Mar 1 '15 at 9:34
• Glad it helped! :) – user220080 Mar 1 '15 at 9:35

I'm not sure why you involve the power set there. You also seemed to have switched from $\in$ to $\subseteq$ somehow when you are talking about null subsets.

Basically, this is a very straightforward application of the assumption, $A\subseteq B$, and the information that $x\notin C$. I will leave it to you to rethink it.

Hint: Consider two cases:

Case 1: $A\subseteq C$ and in this case $A\setminus C=\emptyset$. What can you conclude?

Case 2: $A\setminus C\neq\emptyset$ and choose $x\in A\setminus C$. What can you conclude?

If from $A \subseteq B$ you can deduce that for every set $D$, we have $A\cap D \subset B\cap D$, then:
$$A\setminus C = A\cap C^c \subset B\cap C^c = B\setminus C$$