Is my proof correct? Prove that $A \subseteq B \implies C-B \subseteq C-A$.

Question

Problem: Prove that $A \subseteq B \implies C-B \subseteq C-A$.

My attempt:

$A \subseteq B \implies$ if $x \in A$ then $x \in B$. Since $A$ consists only of some (or all) elements in $B$, we remove less elements of $C$ when taking the set difference, $C - A$, than we do when taking the set difference, $C - B$. Therefore, since for both of these differences we are removing elements from the same set, $C - B \subseteq C-A$.

I appreciate this is pretty sloppy, but I think the logic is almost there, I'm just not sure how to formalise the ideas.

Answer 1

Formally:

$A\subset B$ $\rightarrow $

$B^c\subset A^c$ $\rightarrow$

$C\cap B^c \subset C\cap A^c$.

With

$C - B = C\cap B^c;$ $C-A = C\cap A^c$,

we get:

$C-B \subset C-A.$

Appended:

Proof of $A\subset B$ $\rightarrow$ $B^c\subset A^c$.

$A\subset B $: $x\in A$ then $ x\in B.$

$\rightarrow:$

If $x \not\in B$ then $x \not\in A$, I.e.

$ x\in B^c$ then $x \in A^c$, or

$B^c \subset A^c$.

Answer 2

Element chase.

We want to show $x\in C\setminus B \implies x\in C \setminus A$.

In this context, talking it out will have everything fall into place.

If $x \in C\setminus B$ then $x \in C$ and $x \not \in B$. If $x \not \in B$ then $x \not \in A$. (Why?) So $x \in C$ and $x \not \in A$. So $x\in C \setminus A$. And $ C\setminus B \subset C \setminus A$.

There is a bit of a stickler in that we should prove that if $X \subset Y$ then $x\not \in Y \implies x\not \in X$. The definition of $X \subset Y$ is $x\in X \implies x\in Y$. So the contrapositive of the definition is $x \not \in Y \implies x\not \in X$. That's enough. If it's seems too glib, there is a formal proof by contradiction that:

If $X\subset Y$ and $x \not \in Y$. If $x \in X$ then $x \in Y$ and that is a contradiction so $x\not \in X$.

Answer 3

You are on the right lines, but as you said yourself - the logic isn't quite there. You want to take what you are intuitively saying and turn it into formal logic.

We want to show that $C - B \subseteq C - A$. In other word we want to show that every element of $C - B$ is an element of $C - A$.

So start by showing "If $x \in C - B$, then ..., then $x \in C - A$". Once you have done this you can conclude that $C - B \subseteq C - A$ as $x$ was an arbitrary element.

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As for how you go about filling in "If $x \in C - B$, then ..., then $x \in C - A$", think about what it means for an element to be in $C - B$ and use the fact that $A \subseteq B$.