I proof that, let $A$ a set and $A \neq \emptyset$, then $A \nsubseteq \emptyset$;

Proof by contradiction: if $A \subseteq \emptyset$ then by property I have an absurd , in fact by hypothesis $A \neq \emptyset$, therefore $A \nsubseteq \emptyset$.

Or, direct proof: by hypothesis $A \neq \emptyset$, then "$A \nsubseteq \emptyset$ or $ \emptyset \nsubseteq A$", but by property I have only case that $ A \nsubseteq \emptyset$... Is it correct? Thank you all in advance

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    $\begingroup$ Why if $A\neq\varnothing$ then $A\nsubseteq \varnothing$ or $\varnothing \nsubseteq A$? I think it requires proof, because $\subseteq$ is not total order. (on set of subsets of $X$. $X$ is arbitary set.) $\endgroup$ – Hanul Jeon Mar 16 '13 at 12:04
  • $\begingroup$ Hi @tetori, Ok! Therefore I must prove that $A \nsubseteq \emptyset$, ok? I prove it by contraddiction, so $A \subseteq \emptyset$ then $A = \emptyset$ but by hypothesis $A \neq \emptyset$ and it is contraddiction therefore $A \nsubseteq \emptyset$ Is it correct? $\endgroup$ – mle Mar 16 '13 at 12:19
  • $\begingroup$ I think it is correct :) $\endgroup$ – Hanul Jeon Mar 16 '13 at 12:38
  • $\begingroup$ Hi @tetori... thank you soo much!!! $\endgroup$ – mle Mar 16 '13 at 12:48

In Dijkstra-Feijen style, it is easy to rigorously prove that the two are even equivalent: for any set $A$,

$$ \begin{array}{ll} & A ≠ ∅ \\ \equiv & \;\;\;\text{"expand $≠$ to introduce equality"} \\ & \lnot (A = ∅) \\ \equiv & \;\;\;\text{"introduce double inclusion -- suggested by shape of the desired conclusion"} \\ & \lnot (A ⊆ ∅ \land ∅ ⊆ A) \\ \equiv & \;\;\;\text{"$∅ ⊆ A$ is always true"} \\ & \lnot (A ⊆ ∅) \\ \equiv & \;\;\;\text{"simplify"} \\ & A ⊈ ∅ \\ \end{array} $$

  • $\begingroup$ Wow.. this proof is interesting.. $\endgroup$ – mle Mar 16 '13 at 20:19

Or more positively, if $A\neq\emptyset$ then there exists some $x$ with $x\in A$, and since $x\notin\emptyset$ it follows that $A\not\subseteq\emptyset$.


There is a language issue so I can't quite tell if what you are suggesting is correct or not. In any case, it is not rigorously presented, so I provide you here with a rigorous proof.

Suppose that $A\subseteq \emptyset $. Then, by definition of inclusion, for all $a\in A$ holds that $a\in \emptyset $. However, $a\in \emptyset$ never holds, and thus $a\in A$ never holds, so that $A=\emptyset $. By contrapositive it follows that if $A\ne \emptyset$ then $A$ is not contained in $\emptyset $.

  • $\begingroup$ Hi @Ittay Weiss I corrected my precedent post.... now How is it correct? $\endgroup$ – mle Mar 16 '13 at 11:01

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