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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

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    $\begingroup$ Yes, this also is just fine. $\endgroup$ – Brian M. Scott Mar 14 '13 at 16:15
  • $\begingroup$ Ya thats correct answer $\endgroup$ – kalpeshmpopat Mar 14 '13 at 16:23
  • $\begingroup$ @BrianM.Scott thank you! $\endgroup$ – mle Mar 14 '13 at 18:04
  • $\begingroup$ @kalpeshmpopat thank you! $\endgroup$ – mle Mar 14 '13 at 18:05
  • $\begingroup$ You’re welcome. $\endgroup$ – Brian M. Scott Mar 14 '13 at 18:05
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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 $\lnot\exists \equiv \nexists$)

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  • $\begingroup$ for tomorrow morning $\endgroup$ – mrs Mar 14 '13 at 20:11
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    $\begingroup$ @Oleg: you had all the right ideas and your proof does indeed work. My suggestions are not implying you were wrong: I am simply suggesting how to "tighten up" your proof, and how you can make it a tad more "formal" and precise. $\endgroup$ – Namaste Mar 14 '13 at 20:37
  • $\begingroup$ @amWhy thank you for idea!! $\endgroup$ – mle Mar 14 '13 at 21:02
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Yes, that is a correct proof of your statement by contradiction.

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