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I am having difficulty proving that A-B must be uncountable using proof by contradiction considering (A-B) ∪ B )

Any help?

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If $A-B$ is countable, then $(A-B)\cup B$ is also countable. But $A=(A-B)\cup B$ is uncountable.

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  • $\begingroup$ In general, $A\neq(A-B)\cup B$. But $A\subset(A-B)\cup B$, which is enough. $\endgroup$
    – TonyK
    Sep 18 '16 at 17:44

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