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It is to prove that the empty set is a subset of every set.

Therefore the following proposition is to be proven:

$$ \forall z: ( \emptyset \subset z)~~.$$

Therefore, the converse

$$ \exists z: ( \emptyset \not \subset z)$$

must lead to a contradiction.

Now, considering the set

$$ z = \{\{a\}\} \setminus \{a\} = \{\}~~, $$

if

$$ \emptyset \not \subset z~~, \text{and therefore} ~~ \emptyset \neq z ~~,$$

then $z$ wouldn't be a set, yet it is by definition a set, since the proposition

$$ a \in z $$

is still either true or false. Which is a contradiction.

Assuming my reasoning is flawed, please don't give me the proof. If it is correcct though, I would appreciate an alternative proof. Thank you in advance.

G. Chiusole
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  • Verify with the def of set complement: $B \setminus A = { x \in B \mid x \notin A }$. – Mauro ALLEGRANZA May 25 '17 at 14:26
  • Look at the definition of the empty set along with the definition of set containment. – Dan Brumleve May 25 '17 at 14:26
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    you haven't proved any contradiction, you only proved that the empty set is included in your particular z. But you should be more general and prove that if there exists a z s.t the empty set is not included in z then you have a contradiction. You cannot choose your z. – fonfonx May 25 '17 at 14:32
  • You might want to say $X\subseteq Y$ is defined as the case where $\forall x \in X: x \in Y$. The only possible counterexamples to $\emptyset \subseteq Y$ involve finding an $x$ and $Y$ with $x\in \emptyset$ and $x \not \in Y$; there are no such counterexamples as $\forall x: x \not \in \emptyset$, and so $\emptyset \subseteq Y$ is always true – Henry May 25 '17 at 14:43
  • Recall the definition of a subset: $X\subseteq Y \iff \forall x \in X, x \in Y$, etc... So for all sets A, $\varnothing \subseteq A \iff (x \in \varnothing \rightarrow x \in A).$ This is vacuously true, because an implication with a false antecedent is always true. Since $x \in \varnothing$ is never true (definition of the empty set), the implication $x\in \varnothing \to x \in A$ is true, for any set A we might choose. – amWhy May 25 '17 at 14:43

1 Answers1

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For any $X\subseteq Y$, we have $Y\setminus X\subseteq Y$. So $Y\setminus Y=\varnothing\subseteq Y$.