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I want to simplify $\cup\{A, \cup A\},$ and also $\cup\cup\{A, \cup A\},$ so forth. I thought $\cup\{A, \cup A\}$ would not be simplified more. To say this in plain english, this union is a set that contains 1) all elements in $A$ and 2) all elements in all elements in $A,$ which I am not sure how to simplify. Then $\cup\cup\{A, \cup A\}$ is a set that contains all elements in the first union that I just described. Is it possible to simplify this union more? Thanks for your help.

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  • $\begingroup$ Sorry I didn't see what I posted was included in your post. Yes, I think it is correct and that It cannot be simplified any further. $\endgroup$ – JustDroppedIn Jun 10 at 17:15
  • $\begingroup$ No worries! Thank you very much :) Do you think it'd be possible to simplify the second set? I thought not, but I wanted some more insights. $\endgroup$ – lifeisfun Jun 10 at 17:15
  • $\begingroup$ What kind of notation is this? Is $\cup\{A,\cup A\}$ simply the set $A\cup\bigcup_{x\in A} x$? $\endgroup$ – Evangelos Bampas Jun 10 at 18:05
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    $\begingroup$ @EvangelosBampas : This notation is used when set theory is not merely used in some other areas of mathematics, but rather is done for the purpose of understanding set theory. $$ \underbrace{\quad \cup A \quad}_\text{set theorists' notation} = \underbrace{ \quad \bigcup_{x\in A} x \quad }_\text{common mathematicians' notation} $$ $\endgroup$ – Michael Hardy Jun 10 at 19:17
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    $\begingroup$ @MichaelHardy: Except for the fact that one should still write $\bigcup A$ and not $\cup A$. $\endgroup$ – Asaf Karagila Jun 11 at 9:16
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You might or might not consider it a simplification, but, of course, $\bigcup \left\{A, \bigcup A\right\} = A \cup \bigcup A.$

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