Prove or disprove: Let $S$ be an infinite set such that, $S\cap \mathbb{N}=\emptyset$. Is$|S|=|S\cup \mathbb{N}|$ always true?
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What's $\varphi$? – Oct 26 '19 at 09:34
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1Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Oct 26 '19 at 09:35
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they are disjoint sets @Gae.S. – Azam Iftikhar Oct 26 '19 at 09:36
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2See $X$ is infinite and $Y$ at most countable. Why is $\lvert X \rvert = \lvert X \cup Y \rvert$?. I think this holds without the condition that $S$ and $\mathbb{N}$ are disjoint. – projectilemotion Oct 26 '19 at 09:54
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The set $\Bbb N$ is the smallest infinite set, in the sense that $$ \text{$X$ infinite}\Longrightarrow|\Bbb N|\leq|X|. $$ Thus, if $S$ is infinite one has $|S|=|S\cup\Bbb N|$ regardless if $S$ and $\Bbb N$ are disjoint or not.
Andrea Mori
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1This does not seem to answer the question. How does $|S|=|S\cup\mathbb N|$ follow from the fact that $|\mathbb N|$ is the smallest infinite cardinality? – Andrés E. Caicedo Oct 26 '19 at 21:35
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1@AndrésE.Caicedo: because is a general fact (which I assumed known) that if $A$ and $B$ are sets with $B$ infinite and $|A|\leq|B|$ then $|A\cup B|=|B|$. – Andrea Mori Oct 27 '19 at 08:30