# Differences of sets

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• Feb 21st 2010, 03:02 PM
buddyp450
Differences of sets
If the difference of sets A and B is the set containing elements that are in A but not in B and B is an all-inclusive set then is A-B={null set} or just A-B=null set?

I was thinking that it would be A-B={null set} since the definition of the difference of sets is that it is, "the set containing..."

Thanks!
• Feb 21st 2010, 03:38 PM
Plato
Quote:

Originally Posted by buddyp450
If the difference of sets A and B is the set containing elements that are in A but not in B and B is an all-inclusive set then is A-B={null set} or just A-B=null set?
I was thinking that it would be A-B={null set} since the definition of the difference of sets is that it is, "the set containing..."

I, for one, have no idea what you point is. What are you asking?
Here is a fact: $\displaystyle A \subseteq B\quad \Leftrightarrow \quad A\backslash B = \emptyset$.
• Feb 21st 2010, 04:00 PM
buddyp450
I'm sorry, let me try to be more clear as I'm new to discrete mathematics.

A = {1, 2, 3, 4, 5}
B = {1, 2, 3, 4, 5}

so does

A-B = $\displaystyle \emptyset$

or

A-B = {$\displaystyle \emptyset$}

and could you please explain why?
• Feb 21st 2010, 05:43 PM
simulacrum
Notation
Quote:

Originally Posted by buddyp450
I'm sorry, let me try to be more clear as I'm new to discrete mathematics.

A = {1, 2, 3, 4, 5}
B = {1, 2, 3, 4, 5}

so does

A-B = $\displaystyle \emptyset$

or

A-B = {$\displaystyle \emptyset$}

and could you please explain why?

$\displaystyle \emptyset$ is equivalent to {}. This is standard notation. E.g., {$\displaystyle \emptyset$} is the set containing the empty set. So, A-B is $\displaystyle \emptyset$ or {}, not {$\displaystyle \emptyset$}.
• Feb 21st 2010, 08:39 PM
Drexel28
Quote:

Originally Posted by Plato
I, for one, have no idea what you point is. What are you asking?
Here is a fact: $\displaystyle A \subseteq B\quad \Leftrightarrow \quad B\backslash A = \emptyset$.

Is that a typo, or am I misunderstanding the notation? $\displaystyle \{1\}\subset\{1,2\}$ but $\displaystyle \{1,2\}-\{1\}=\{2\}$
• Feb 22nd 2010, 03:02 AM
HallsofIvy
I think that by "B is an all inclusive set" you mean that B is the universal set. In that case, everything is in B so "are in A but not in B" does not apply to anything since there is nothing that is "not in B". A- B is the empty set.