# Duplication in sets

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.

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

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 $B$.

• 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. Mar 8, 2015 at 16:01
• It has to hold for all sets simultaneously. Mar 8, 2015 at 16:08
• 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. Mar 8, 2015 at 16:09

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