I am doing a text that my big brother gave me: Mathematical Analysis I - Zorich. This stuff is pretty hard for me, since in class we don't do sets.

I can see why they are true with pictures, but i don't know how to prove them mathematically:

$A,B,C\subset M$

1.a) $(A \subset C) \wedge (B \subset C) \Leftrightarrow ((A \cup B) \subset C)$

I can see that if the left is true, the right is true, and if one is false, the other is false.

but what if $A$ exclusively OR $B$ subsets C, then the right is true, but the left is false?

1.b) $(C\subset A) \wedge (C \subset B) \Leftrightarrow (C\subset (A \cap B))$

This seems obvious, but how to mathematically show it? i can just draw the Venn diagram again, but that is not math?

Side question: Do Venn diagrams count as mathematical proofs? Or are they just a tool to clear things up?


If $A$ exclusively or $B$ are contained in $C$, then $A\cup B$ is not a subset of $C$. Without loss of generality, we can assume that $A\subset C$ and $B\not\subset C$. If $A\cup B\subset C$ then $B\subset C$.

I'll give the hint of 1.b): use the definition of intersection $$A\cap B=\{x:x\in A\land x\in B\}.$$

Let assume $x\in C$ then $x\in A$ and $x\in B$ so...

  • $\begingroup$ is my attempt to solve 1b from your answers advice correct(see second answer) $\endgroup$ – beginner Nov 24 '14 at 1:52
  • $\begingroup$ @beginner So, your question is using Venn diagrams as mathematical proof is valid? $\endgroup$ – Hanul Jeon Nov 24 '14 at 1:54
  • $\begingroup$ that was a secondary question. was my attempt below correct by the way? $\endgroup$ – beginner Nov 24 '14 at 2:09
  • $\begingroup$ @beginner I think it has some error. $C\subset A$ and $C\subset B$ does not imply $C\subset A\cap B$ directly. But, since if $x\in C$ then $x\in A$ and $x\in B$, so $x\in A\cap B$ by definition of $A\cap B$ so $x\in C\implies x\in A\cap B$ (that is, $C\subset A\cap B$.) $\endgroup$ – Hanul Jeon Nov 24 '14 at 2:12
  • $\begingroup$ is that better? thank you, I think i get it now! $\endgroup$ – beginner Nov 24 '14 at 2:25

$$(C\subset A) \wedge (C \subset B) \Leftrightarrow (C\subset (A \cap B))$$

Proof: $$(C\subset A) \wedge (C \subset B) \implies (C\subset (A \cap B))$$ Assume $C\subset A$ and $C\subset B$. This means $\forall x\in C, x\in A$ and $x\in B$. From this we can see that $\forall x\in C,x\in (A\cap B)$. Since this is all $x\in C$, $C\subset A\cap B$.

We must now prove that:$$(C\subset A) \wedge (C \subset B) \Longleftarrow (C\subset (A \cap B))$$

Assume $C\subset (A\cap B)$ hence $\forall x\in C, x\in A\cap B$, since $(A\cap B) \subset A$ and $(A\cap B) \subset B$, $\forall x\in C,x\in A$ and $x\in B$

thus it is proved.


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