I'm working on a problem that involves the cardinality of a set and its complement. Basically, we have a set $A$ and its complement $A^c$ and I want to know if the conjecture $|A \cup A^c| > |A^c|$ or $|A \cup A^c| \geq |A^c|$ is true. A formal theorem or proof would be much appreciated.
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2If you consider the complement of a set $A$, you need a universal set $U$ containing $A$. Then the complement is $U\setminus A$. So $A\cup A^c = U$ and this answers your question. – Wuestenfux Aug 23 '19 at 14:28
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In any universe $U$ the complement of $A$ is a subset of $U$. That should let you prove the weak inequality and disprove the strict inequality, – Ethan Bolker Aug 23 '19 at 14:31
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In your title, I assume that its $A \cup A^c$ ? – Jean Marie Aug 23 '19 at 14:43
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If A is empty set , the strict inequality is wrong , if complement of A is empty set , the weak inequality is true. – nissim abehcera Aug 23 '19 at 14:54
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Fix a universal set $X$ and a subset $A\subseteq X$. Since $A^c\subseteq A\cup A^c$ (in fact, $A\cup A^c=X$), we do indeed have $|A\cup A^c|\geq |A^c|$.
However, a strict inequality does not always hold. Indeed, consider $X:=\mathbb R$ and $A:=\mathbb R\setminus [0,1]$, so we have $|A\cup A^c|=|\mathbb R|=|[0,1]|=|A^c|$.
Dave
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