I don't understand why this theorem is false.

Suppose that $A \subseteq C$, $B \subseteq C$, and $x \in A$. Then $x \in B$.

Invalid Proof:

Suppose that $x \notin B$. Since $x \in A$ and $A \subseteq C$, $x \in C$. Since $x \notin B$ and $B \subset C$, $x \notin C$. But now we have proven both $x \in C$ and $x \notin C$, so we have reached a contradiction. Therefore $x \in B$.

I'm thinking that $$\forall x(x \in B \implies x \in C) = \forall x(x \notin B \vee x \in C)$$

It's true that $x \notin B$, so this statement is true, and can't be used to prove through contradiction that $x \in B$. But I'm not completely sure, so any clarification would help here. Thanks.

  • $\begingroup$ Welcome to Mathematics Stack Exchange. Here is a MathJax tutorial. $x\not\in B$ and $B\subseteq C $ does not imply $x\not\in C$ $\endgroup$ – J. W. Tanner Sep 4 '19 at 22:49
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    $\begingroup$ A simple countra example is disjoint A and B. $\endgroup$ – William Elliot Sep 4 '19 at 23:07
  • $\begingroup$ Try $C$ the set of all integers, $A$ the set of even integers, and $B$ the set of odd integers. Always try to play with concrete examples. Sometimes pictures (like Venn diagrams, in this situation) are instructive. $\endgroup$ – Ted Shifrin Sep 4 '19 at 23:18
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    $\begingroup$ "Since $x \not \in B$ and $B\subset C$, $x \not \in C$". This is not valid. $B\subset C$ so all the elements in $B$ are in $C$ but not all the elements in $C$ are in $B$. ANd if an element is not in $B$ there is utterly no reason it can't be one of the elements in $C$ that aren't in $B$. Consider: Penguins $\subset$ Birds. Now Daffy Duck $\not \in $ Penguins. And Penguins $\subset$ Birds. So you are claiming that therefore Daffy Duck $\not \in $ Birds. But Daffy Duck could be (and is) a bird that is not a penguin. $\endgroup$ – fleablood Sep 4 '19 at 23:22
  • $\begingroup$ You got the formula right $\endgroup$ – Shaun Sep 4 '19 at 23:47

"since $x \notin$ B and $B \subseteq C, x \notin C$" That's the mistake. that implication is wrong. Remember the definition of subsets. If $B \subseteq C$, this means that for all elements in $B$, they are also in $C$. Or with symbols $x \in B \rightarrow x\in C$. this statement says nothing about $x$ not being in $B$, so if an element is not in $B$, you can't deduce anything. that's why that implication is wrong. Look, here's a simple example. Let $C = \{w, x, y, z\}$ and $B = \{x, y\}$, so $B$ is a subset of $C$. The element $z$ is not in $B$ but it is in $C$.

  • $\begingroup$ It might be easier to understand this if in the example at the end you use $C$ rather than $A$ and also use different letters for the set elements $\endgroup$ – J. W. Tanner Sep 4 '19 at 23:30

Since $x \not\in B$ and $B \subseteq C$, $x \not\in C$.

Actually, if $x \not\in B$ and $B \subseteq \color{blue}{C}$, both $\color{red}{x \in C}$ and $\color{magenta}{x \not\in C}$ are possible. In a picture:

enter image description here


Here's a counter model . . .

Let $A=\{1\}, B=\{2\}, C=\{1,2\}$. Then $A\subseteq C$ and $B\subseteq C$ with $1\in A$ but $1\notin B$.



$C= \{1,2,3,4,5,6,7,8,9,10,11,12\}$, all the integers up to $12$.

And $A = \{2,4,6,8,10,12\}$, all the even numbers up to $12$.

And $B = \{1,4,9\}$, all the perfect squares up to $12$.

We note that $A\subset C$ and $B\subset C$.

That theorem says that if $x\in A$ then $x \in B$.

This is clearly not true. We have $2,6,8,10,12$ all in $A$ yet none of them is in $B$.

Let's assume $x \in A$. So $x = 2,4,6,8,10,$ or $12$.

Let's go to the proof.

Suppose $x \not \in B$

(okay, so $x \ne 1,4,9$ in particular $x \ne 4$ but $x$ could be $2,6,8,10,12$ still.)

Since $x \in A$ and $A \subset C$, $x \in C$.

(This follows. If $x=2,6,8,10$ or $12$ we do have $x \in C$.)

Since $x \not \in B$ and $B\subset C$ then $x \not \in C$.

Does that actually make sense? $x\ne 1,4,9$ and $1,4,9\in C$ but there are other things in $C$ as well other than those. We could have $x=2,6,8,10$ or $12$.

There is a concept of $C\setminus B = \{x\in C$ but where $x \not \in B\}$. As we know that $x\not \in B$ that if $x\in C$ then we must have $x\in C\setminus B$. But $C\setminus B$ certainly doesn't have to be empty!

So that doesn't follow at all!

But now we have proven both $x ∈ C$ and $x \not \in C$.so we have reached a contradiction. Therefore $x ∈ B$.

Except we haven't proven $x \not \in C$. We have reached no contradiction. And $x$ need not be in $B$. $x\in C$ so $x$ could be in $C \cap B$ or $x$ could be in $C\setminus B$. But we have no way of telling.


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