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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)?

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    $\begingroup$ What do you mean by $A \in B$ ? Is A a set or an element? $\endgroup$ – Collapse Feb 4 '17 at 19:42
  • $\begingroup$ @DomoB the question in my book says that all three are sets $\endgroup$ – user384262 Feb 4 '17 at 19:44
  • $\begingroup$ 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. $\endgroup$ – Carsten S Feb 4 '17 at 19:44
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    $\begingroup$ Then what is the difference between $\in$ and $\subseteq$? $\endgroup$ – Alberto Feb 4 '17 at 19:44
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    $\begingroup$ If $A \in B$, and $A$ is a set, then $B$ is a set of sets, like $\{\{1\},\{2,3\}\}$. $\endgroup$ – Fabio Somenzi Feb 4 '17 at 19:45
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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 $\subset$).

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

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