TITLE: Proof by contradiction: $ \emptyset \subseteq A$
QUESTION [5 upvotes]: I have to proof by contradiction that: let $ A $ a set and $ \emptyset $ the empty set, then $ \emptyset \subseteq A$;  if  $ \emptyset \nsubseteq A$ then  $\exists x \in \emptyset  ( x \notin A ) $ but for hypothesis "let $ \emptyset $ the empty set, then $\nexists x \in \emptyset$", so I have a contradiction and therefore $ \emptyset \subseteq A$ is true! Is it correct? Thank you all in advance

REPLY [9 votes]: Nitpick (very slight alteration to follow):

Let $ A $ [be] a set and $\emptyset$ the empty set. Then $ \emptyset \subseteq A$.  

Proof:
[Let $A$ be a set and $\emptyset$ the empty set. Suppose also, for the sake of contradiction, that] $\; \emptyset \nsubseteq A$.
Then $\exists x \in \emptyset,$ [such that] $( x \notin A ) $.
But by hypothesis, $\emptyset$ is the empty set, [thus by the definition of the empty set], $\lnot\exists x \in \emptyset$.
So [we] have [reached] a contradiction, and it must therefore follow that] $ \;\;\emptyset \subseteq A,$ [as desired].

(Note: here $\lnot\exists \equiv \nexists$)