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I am trying to decipher the difference between the following two statements:

If $ A \in B $ and $ B \subseteq C $ then $ A \in C $.

vs.

If $ A \in B $ and $ B \subseteq C $ then $ A \subseteq C $.

I think the first statement is the true statement and the 2nd is false because of $ A \in B $ not $A \subseteq B $

Am I on the right track? How would I give a simple counter example to show the statement is false?

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  • $\begingroup$ You are on the right track. The first is, in some places, the definition of $B\subseteq C$. $\endgroup$
    – Thomas Andrews
    Commented Feb 25, 2015 at 18:51
  • $\begingroup$ see also here $\endgroup$
    – user 1
    Commented Feb 25, 2015 at 19:53
  • $\begingroup$ @Alyssa Wallace: please mark "accept" for one answer (that you think is the best) and exclude the question from "unaccepted list" $\endgroup$
    – user 1
    Commented Feb 25, 2015 at 20:02

3 Answers 3

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Another example. These are true $$ -5 \in \mathbb Z\quad\text{and}\quad \mathbb Z \subseteq \mathbb Q $$ and it follows that $$ -5 \in \mathbb Q $$ but it does not follow that $$ -5 \subseteq \mathbb Q $$

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That's right. That is pretty much the definition of a subset.

For the second statement consider

$A:=\{1\}$, $B:=\{\{1\},2\}$ and $C:=\{\{1\},2,3\}$.

Then $A\in B$, $B\subseteq C$, but $A\not\subseteq C$, since $1\notin C$.

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  • $\begingroup$ Does the notation of the brackets make a difference? Can you just have $ B $ be the set $ \{1, 2\} $ and $ C $ be the set $ \{1,2,3\} $ $\endgroup$
    – user219081
    Commented Feb 25, 2015 at 19:03
  • $\begingroup$ @Alyssa Wallace Sure but then $ A $ isn't a set so $ A\subseteq C $ makes no sense. $\endgroup$
    – Uncountable
    Commented Feb 25, 2015 at 19:09
  • $\begingroup$ @Uncountable as commented above $A=1$ is still a set under usual mathematical assumptions. $\endgroup$
    – DRF
    Commented Feb 26, 2015 at 7:01
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First statement:

if A is an element in B, and B is a subset of C, then A is in C.

Explanation: this is true because B being a subset of C means that every element in B is also in C.

Second statement:

if A is an element in B and B is a subset of C, then A is a subset of C.

Explanation: this is false because the element A is not necessarily a set. For example, let $C = \{1,2,3\}$. Let $B = \{1,2\}$. Let $A$ be $1$. $1$ is not a subset of $\{1,2,3\}$ -- it is a member.

Hypothetically: $\{1\}$, on the other hand, would be a subset.

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  • $\begingroup$ If you subscribe to modern foundation theory just about everything is a set. The example you give is correct but under usual assumptions if $A=1$, $B=\{1,2\}$ and $C=\{0,1,2\}$ then $A\in B$ $B\subseteq C$ and $A\subseteq C$ since if you parse what $A$ is given the Von Neuman definition of ordinals you end up having $A=\{\{\}\}$ and $0=\{\}$. $\endgroup$
    – DRF
    Commented Feb 26, 2015 at 6:57

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