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To prove:

If $A \triangle B=A$, then $ B= \emptyset$.

This seems simple enough, as an idea. I mean if the set $B$ is anything but empty, $A\triangle B$ would contain more or less than simply $A$, simple examples would be two disjoint sets or two intersecting sets or whenever $A=B$. But how do I prove it formally?

The example I gave feels like I can make it work with a contrapositive: If $B$ is not empty, $A\triangle B$ does not have the same elements as $A$.

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    $\begingroup$ You should know that $A \Delta A = \emptyset$. Then $$A \Delta A \Delta B =A \Delta A $$ $$\emptyset \Delta B= \emptyset$$ $$B= \emptyset$$ $\endgroup$ – Crostul Sep 5 '18 at 16:11
  • $\begingroup$ That's brilliant. $\endgroup$ – Wesley Strik Sep 5 '18 at 18:14
  • $\begingroup$ Nice for a direct proof :) $\endgroup$ – Wesley Strik Sep 5 '18 at 18:24
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My proof is by contradiction.

Let us recall that the symmetric difference is defined as $$A \triangle B = (A\setminus B)\cup(B\setminus A)$$

Thus, if $B$ was not empty, it would contain an element $b$. I distinguish two cases:

  1. If $b$ also lies in $A$, then by hypothesis it lies in $A \triangle B=A$. This is absurd, because $A \triangle B$ does not contain any element of $A\cap B$.
  2. If $b$ does not lie in $A$, then it lies in $B\setminus A$. Hence, by definition, it must lie in $A \triangle B$. But by hypothesis, this is no other than $A$, which leads us to an absurdity.

We conclude that $B$ must be empty.

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  • $\begingroup$ This is one pretty proof! $\endgroup$ – Wesley Strik Sep 5 '18 at 18:23
  • $\begingroup$ Stuff like this makes me appreciate and fall in love with mathematics every single day. $\endgroup$ – Wesley Strik Sep 5 '18 at 18:24

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