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Thread: Show that if A and B are sets with A ⊆ B, then A ∪ B = B.

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    Show that if A and B are sets with A ⊆ B, then A ∪ B = B.

    Show that if A and B are sets with A ⊆ B, then A ∪ B = B.

    Is A ∪ B = B. because A ∪ B, Means A or B or Both ?
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    Re: Show that if A and B are sets with A ⊆ B, then A ∪ B = B.

    Quote Originally Posted by lamentofking View Post
    Show that if A and B are sets with A ⊆ B, then A ∪ B = B.
    Is A ∪ B = B. because A ∪ B, Means A or B or Both ?
    Basically, yes.
    Realize, That if $\displaystyle X$ is a set and $\displaystyle Y$ is any other set then $\displaystyle X\subseteq (X\cup Y)$.

    So in the case $\displaystyle B\subseteq (A\cup B)$. Now all you need do is show $\displaystyle A\cup B\subseteq B.$
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    Re: Show that if A and B are sets with A ⊆ B, then A ∪ B = B.

    Quote Originally Posted by Plato View Post
    Now all you need do is show $\displaystyle A\cup B\subseteq B.$
    Is the answer because A U B means that A or B so it can be included in B ?
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    Re: Show that if A and B are sets with A ⊆ B, then A ∪ B = B.

    Quote Originally Posted by lamentofking View Post
    Is the answer because A U B means that A or B so it can be included in B ?
    Yes.
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    Re: Show that if A and B are sets with A ⊆ B, then A ∪ B = B.

    The rigorous way of showing that "$\displaystyle X= Y$" is to show that "$\displaystyle X\subseteq Y$" and that "$\displaystyle Y\subseteq X$". And to show that "$\displaystyle X\subseteq Y$" start with "if $\displaystyle x\in X$ and use the conditions on X and Y to conclude "therefore $\displaystyle x\in Y$".

    So to show that, $\displaystyle A\cup B= B$ you must first show $\displaystyle A\cup B\subseteq B$. And you do that by saying "if $\displaystyle x\in A\cup B$ then either $\displaystyle x\in A$ or $\displaystyle x\in B$ (by definition of "$\displaystyle A\cup B$" and then do it in two cases:
    1) If $\displaystyle x\in B$ we are done.

    2) if $\displaystyle x\in A$ then because $\displaystyle A\subseteq B$, $\displaystyle x\in B$.

    So that in either case, if $\displaystyle x\in A\cup B$ then $\displaystyle x\in B$.

    All that remains is to show that $\displaystyle B\subseteq A\cup B$. To do that, if $\displaystyle x\in B$ then $\displaystyle x\in A\cup B$ by definition of $\displaystyle A\cup B$.
    Last edited by HallsofIvy; Oct 6th 2013 at 05:02 PM.
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