Let $A$ be a infinite set and let $D\subseteq A$ be a numerable set so that $A-D$ in infinite. Prove that $(A-D)\sim A$.
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You don't say what $A$ is a subset of, so I'll prove this in a general setting. I'll also assume you mean countable by "numerable".
If $A$ is countable then the result is clear: $A - D \subset A$ is an infinite subset of a countable set, so it is also countable.
To prove the result for uncountable $A$, assume $\left|A - D\right| < \left|A\right|$. The union $D \cup (A - D)$ is equal to $A$, but its cardinality is $\left|A - D\right|$ because $D$ is countable and $A - D$ is infinite. This is a contradiction. We conclude that $(A - D) \sim A$.
Ayman Hourieh
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I see what you were saying. My post was likely more confusing than helpful. I missed the ...so $A - D$ is infinite... – amWhy Mar 27 '13 at 02:13
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Ayman How to prove that $|D\cup (A-D)|=|A-D|$? – Roiner Segura Cubero Mar 27 '13 at 06:02
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@RoinerSeguraCubero See this answer. If it doesn't make sense, tell me what you know, in what context you countered this problem, etc, and I'll try to help further. – Ayman Hourieh Mar 27 '13 at 21:15