# elementary set theory (being a member and subset of a set)?

I have to prove or give a counterexample for these two statements:

For the following statements about sets $A$, $B$, and $C$, either prove the statement is true or give a counterexample to show that it is false.

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

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

I tried to do it by creating random sets like $A = \{2\}$, $B = \{2,3\}$ and $C = \{2,3,4\}$ and so both statements would be true right? I can't think of a counterexample but I don't know how to actually prove these statements.

Also, if A was the empty set then wouldn't both statements always be true (because the empty set is a member of every other set)?

• What do you mean by $A \in B$ ? Is A a set or an element? – Collapse Feb 4 '17 at 19:42
• @DomoB the question in my book says that all three are sets – user384262 Feb 4 '17 at 19:44
• The statements whose truth they want you decided are: “For all sets $A$, $B$, $C$ such that $A\in B$ and $B\subseteq$ it is true that $A\subseteq C$.” and similarly for the second. – Carsten S Feb 4 '17 at 19:44
• Then what is the difference between $\in$ and $\subseteq$? – Alberto Feb 4 '17 at 19:44
• If $A \in B$, and $A$ is a set, then $B$ is a set of sets, like $\{\{1\},\{2,3\}\}$. – Fabio Somenzi Feb 4 '17 at 19:45

This first statement is false.

A counter-example could be: $A=\{1\}$, $B=\{\{1\}\}$ and $C=\{\{1\},\{2\}\}$.

Then you have $A\in B$ and $B\subset C$, but you don't have $A\subset C$.

The second statement is true.

To prove it, take $A$, $B$ and $C$ meeting all the conditions. Then $A\in B\subset C$, so you do have $A\in C$ (by the definition of $\subset$).

• $A$ should be a set – Alberto Feb 4 '17 at 19:47
• I wouldn'y say it's by transitivity, but rather it is simply the definition of $\subset$. – Ittay Weiss Feb 4 '17 at 19:49
• @Alberto I edited to take account of your remark. – E. Joseph Feb 4 '17 at 19:50
• @IttayWeiss May be that is a simpler argument indeed :) – E. Joseph Feb 4 '17 at 19:51
• I meant to be helpful. Perhaps the tone was not nice. Anyway, I'll remove my comment. Thanks for the edit! – Namaste Feb 4 '17 at 22:12