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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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    What do you mean by $A \in B$ ? Is A a set or an element? – 2017-02-04
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    @DomoB the question in my book says that all three are sets – 2017-02-04
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    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. – 2017-02-04
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    Then what is the difference between $\in$ and $\subseteq$? – 2017-02-04
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    If $A \in B$, and $A$ is a set, then $B$ is a set of sets, like $\{\{1\},\{2,3\}\}$. – 2017-02-04
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    @DomoB It can be both. The question, as it is right now, is perfectly logical and correct. – 2017-02-04
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    In using $\in$, e.g. $x\in y$, the only requirement is that **$y$** is a set; **$x$** can be anything at all. – 2017-02-04
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    @amWhy thanks!! – 2017-02-04
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    @123 Just a side comment: you cannot prove a statement about some relation holding for all A, B, C, by showing only that a specific example of the relation holds. However, you *can* disprove a universal statement about A, B, C, by providing a counter-example of sets A, B, C, for which the statement fails to hold. – 2017-02-04

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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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    $A$ should be a set – 2017-02-04
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    I wouldn'y say it's by transitivity, but rather it is simply the definition of $\subset$. – 2017-02-04
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    @Alberto I edited to take account of your remark. – 2017-02-04
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    @IttayWeiss May be that is a simpler argument indeed :) – 2017-02-04
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    I meant to be helpful. Perhaps the tone was not nice. Anyway, I'll remove my comment. Thanks for the edit! – 2017-02-04
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    Glad to have helped! – 2017-02-04