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I have two questions.

$1.$ From what i know the sets $A = \{1,1,2\}$ and $B = \{1,2\}$ are the same set. So my first question is, is $A\cup A=A$, or even further, is the union of a set and any of its subsets still that set? Say, for example $B \subseteq A$. Is $A\cup B=A$ as $A\cup B = \{1,1,1,2,2\} = \{1,1,2\} = \{1,2\}$.

$2.$ If the first question has a positive answer then how is the following theorem possible: Theorem: There is a unique set $A$ such that for every set $B$, $A \cup B = B$.

Proof: Existence: Clearly $\forall B(\emptyset \cup B = B)$, so $\emptyset$ has the required property. Uniqueness: Suppose $\forall B(C \cup B = B)$ and $\forall B(D \cup B = B)$. Applying the first of these assumptions to $D$ we see that $C \cup D = D$, and applying the second to $C$ we get $D \cup C = C$. But clearly $C \cup D = D \cup C$, so $C = D$.

EDIT: since I guess this is a vague question what I am asking is: How is it possible that according to the theorem there is one UNIQUE set(the empty set in the proof) for which $A \cup B = B$. If the answer to the first question is positive than there are more than one sets (any subset of $B$ or $B$ itself) for which the equality $A \cup B = B$ is true.

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    $\begingroup$ The uniqueness comes in when it is demanded that $A\cup B=B$ for every set $B$. Not just a single set $B$. $\endgroup$
    – drhab
    Mar 8, 2015 at 16:14
  • $\begingroup$ Yes that is correct. Thank you. $\endgroup$
    – alexgiorev
    Mar 8, 2015 at 16:16

3 Answers 3

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  1. It is true in every case that $A\cup A=A$, and it is true that if $B\subseteq A$ then $A\cup B=A$.
  2. The set $\emptyset$ is called "the empty set." It is the set that doesn't contain any elements (hence the word "empty" it its name).

If $A$ and $B$ are sets, then $A\cup B$ as the smallest set such that both $A\subseteq (A\cup B)$ and $B\subseteq (A\cup B)$ are true. Bearing in mind that the empty set is a subset of every set, we have $\emptyset\cup B=B$ because $B$ is the smallest set such that $B\subseteq B$ and $\emptyset\subseteq B$ are both true.

Note that no set (besides the empty set) is a subset of every set. Therefore, the emptyset is the only set $A$ such that $A\cup B=B$ for every set $B$.

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  • $\begingroup$ The theorem states that there is a unique set A for which A ∪ B = B. Clearly A, any subset of A and ∅ are a couple of possibilities. Therefore there is no UNIQUE set, as there is more than one. $\endgroup$
    – alexgiorev
    Mar 8, 2015 at 16:01
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    $\begingroup$ It has to hold for all sets simultaneously. $\endgroup$ Mar 8, 2015 at 16:08
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    $\begingroup$ The key to this theorem is that $\emptyset\cup B=B$ for any set $B$. It is true that if we pick a particular set $B$ then there could be a good number of sets $A$ such that $A\cup B=B$. However, the empty set is the only set $A$ such that $A\cup B=B$ regardless of which set $B$ we choose. $\endgroup$
    – Jasha
    Mar 8, 2015 at 16:09
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The answer to your first question is positive. Consequently $A$$\cup B=A$ for each subset $B$ of $A$.

The empty set is a subset of every set $A$ so that $A\cup\varnothing=A$ for every set $A$.

If $A\cup B=A$ is true for each $A$ then it is also true for $A=\varnothing$ leading to $B=\varnothing\cup B=\varnothing$.

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  • $\begingroup$ @user1 In the questions 1) and 2) of the OP the role of set $A$ is somehow switched. I don't follow that inconsistency in my answer. So it could be vague in the sense that is does not seem to match the question. However, it proves clearly the mentioned uniqueness. $\endgroup$
    – drhab
    Mar 8, 2015 at 16:25
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  1. Repetition (and order) don't change a set; see here, for example.
  2. your proof is correct. in other words: $$A\cup B = B , \forall B \iff A\subset B , \forall B \iff A = \emptyset$$
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  • $\begingroup$ Downvote? Well, be so kind to explain why. $\endgroup$
    – user 1
    Mar 8, 2015 at 16:17

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