# $A \subset S$ then $A\cap S =A$

How would I prove the following $$A \subset S$$ then $$A\cap S =A$$

Not sure I been thinking of an example say $$A={1,2,3}$$ and $$S={1,2,3,4,5,6}$$

So I tried the following $$(x:x\in A \wedge x\in S)$$

so I let $$x:x\in A$$ I am not sure if this finishes the proof. Because I can can also $$x\in S$$ I think it would also be true if $$(x:x\in S)$$

• Why not just show a two sided inclusion?...
– Mark
Dec 17, 2021 at 22:01
• I think for the second part I can do $x\in A$ then because A is subset of S then $x\in S$ hence $x\in S \wedge A$ Dec 17, 2021 at 22:18

We can prove the equality of two sets, $$X=Y$$ by showing $$Y-X=\Phi$$ and $$X-Y=\Phi$$. Let's try the same here.

(1) Prove that $$A-(A\cap S)=\Phi$$

$$A-(A\cap S)$$

$$=\{x:x\in A-(A\cap S)\}$$

$$\Leftrightarrow \{x:x\in A \wedge x \notin A \cap S\}$$

$$\Leftrightarrow \{x:x\in A \wedge (x\notin A \lor x\notin S)\}$$

$$\Leftrightarrow \{x:(x\in A \wedge x\notin A) \lor (x\in A \wedge x\notin S)\}$$, by the distributivity of $$\wedge$$ over $$\lor$$

$$\Leftrightarrow \{x:(x\in A \wedge x\notin A) \lor (x\in S \wedge x\notin S)\}$$, since by $$A \subset S$$, $$x \in A \implies x \in S$$

$$=\Phi$$

(2) Prove that $$(A\cap S)-A=\Phi$$, this is trivial (a tautology for any sets $$A$$, $$S$$)

$$(A \cap S) - A$$

$$=\{x:x\in (A \cap S) - A\}$$

$$\Leftrightarrow \{x:x\in A \cap S\wedge x\notin A\}$$

$$\Leftrightarrow \{x:(x\in A \wedge x \in S) \wedge x\notin A\}$$

$$\Leftrightarrow \{x:(x\in A \wedge x\notin A) \wedge x \in S\}$$, by commutativity and associativity of $$\wedge$$

$$=\Phi$$

Combining (1) and (2), we have $$A=A\cap S$$

• excellent proof sir Dec 18, 2021 at 2:16

I would say it's pretty easy to show this through a Venn diagram, if it helps you.

Also, if you have a set S for which $$A \subset S$$, then any $$a\in A$$ is also in S, but not every $$s\in S$$ is in A. So, you have:

$$A\cap S = \{s\in S | s \in A\} = \{s \in A\} = A$$

I think that's enough if you just want to understand why that is, but for more detail look here: https://proofwiki.org/wiki/Intersection_with_Subset_is_Subset

To prove that two sets are equal, you prove that they have the same elements. The elements of $$A$$ are, well, the elements of $$A$$. The elements of $$A \cap S$$ are, by definition, the elements of $$A$$ which are also elements of $$S$$. But all of them are elements of $$S$$, since that is precisely what $$A \subseteq S$$ means.