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let $A\;, B$ and $C$ be subsets of $E$. show that

$$A\cup B =E \iff \overline A\subset B$$

Notation :

  • Absolute complement of $A$: $\overline A=\mathcal{C}_{E}^{A}=\{ x\in E \mid x\notin A \}$

My thoughts

  • Method $1$ : Via Logic theory

$$A\cup B =E \iff \overline A\subset B$$ we can see that as Tautology in logic theory and we can prove it by truth table: $$(A\vee B)\iff ((\neg A)\to B)$$

\begin{array} {|c|} \hline A & B & (A\vee B )& \neg A & (\neg A \to B )& [(A\vee B)\iff ((\neg A)\to B) ]\\ \hline T & T & T & F & T & T \\ \hline T & F & T & F & T & T\\ \hline F & T & T & T & T & T\\ \hline F & F & F & T & F & T\\ \hline \end{array}

  • Method $2$ : Via Venn diagram ( Venn diagram not rigorous so they are not really proofs )

enter image description here

I think I have two possibilities to draw Venn diagram for $A\cup B =E \implies \overline A\subset B$

$1$. $B=\overline A\qquad $ ($A\cup B=A\cup \overline A =E \implies \overline A\subset B$ )

$2$. $B=E\qquad $ ($A\cup B=A\cup E =E \implies \overline A\subset B$ )

  • Method $3$

    1. First step : show that : $\overline A\subset B \implies A\cup B =E$

Suppose that $\overline A\subset B$ and let's prove that $ A\cup B =E$

  • we have $A \subset E$ and $B \subset E$ then $$(A\cup B) \subset E\quad (1)$$
  • Conversely, let $x\in E $ we have two cases $x\in B $ or $x\notin B$

First case: if $x\in B$ then $x\in A\cup B $

Second case: if $x\notin B$ then $x\notin \overline A\quad (\text{since}\; \overline A \subset B )$ thus $x\in A $

Therefore $x\in A\cup B $ so in both cases we have : $$x\in E\implies x\in A\cup B$$ Therefore $$E\subset (A\cup B)\quad (2) $$ From $(1)$ and $(2)$ we conclude that $$E=A\cup B$$

  1. Second step : show that : $A\cup B =E \implies \overline A\subset B$

Suppose that $A\cup B=E$ and and let's prove that $\overline A\subset B$ Let $x\in \overline A$ we have: \begin{align*} x\in \overline A &\implies x\in E \\ &\implies x\in A\cup B \\ &\implies x\in B (\;\text{since}\; x\notin A )\\ x\in \overline A &\implies x\in B \end{align*} Therefore $$A\cup B =E \implies \overline A\subset B$$

Conclusion:

$$A\cup B =E \iff \overline A\subset B$$


  • Is my proof correct also I'm interested in more ways of proving it
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  • $\begingroup$ The relative complement $\overline{A}$ in OP is usually written as $E\setminus A$. $\endgroup$
    – user9464
    Dec 11, 2015 at 15:57
  • $\begingroup$ it's Absolute complement $\overline A $ $\endgroup$
    – Educ
    Dec 11, 2015 at 16:06
  • $\begingroup$ No reason why Venn diagrams are not rigorous. With a little care in the definitions and axioms to prevent hidden assumptions. $\endgroup$ Dec 12, 2015 at 9:04
  • $\begingroup$ this is why math.stackexchange.com/questions/304173/… $\endgroup$
    – Educ
    Dec 12, 2015 at 9:22
  • 1
    $\begingroup$ for instance blog.stevemould.com/venn-vs-euler-diagrams $\endgroup$
    – Leox
    Dec 17, 2015 at 19:44

2 Answers 2

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  • if $x\in \bar{A}$, then $x\not \in A$, therefor by $A\cup B=E$, we have $x\in B$, so $\bar{A}\subseteq B$.

  • If $\bar{A} \subseteq B$, then $A\cup \bar{A}\subseteq A\cup B$, so $E\subseteq A\cup B$, therefore $A\cup B =E$.

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This is how I would do it:

$A\cup B=E\;$ then $\; E/A \subseteq A \cup B\;$ and then $\;E/A \subseteq B\;$ because $\;E/A\;$ isn't a a subset of $A$ by definition.

And if:$\quad E/A \subseteq B\;$ then $\; A\cup E/A\subseteq A\cup B\;$. Hence $A\cup B=E$

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