Set Operations: Philosophy Forums

Set Operations

Christim
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Posted 10/02/09 - 10:58 AM: Subject: Set Operations	#1
2) True or false? a) EMPTYSET MEMBER {EMPTYSET} b) EMPTYSET MEMBER {0, {EMPTYSET}} c) {0} SUBSET { {0} } d) {{a},{b},c} SUBSET {a,b,{c}} e) {3,{4}} SUBSET {3,{3,{4}}} I am confused right now. I need some help. How can you tell something is a member or subset of something? What is the difference between member and subset?

rigelrover
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Posted 10/02/09 - 11:03 AM:	#2
In some cases a subset can be a single member. A subset is a set of members that all belong to the original set. a) false b) true (but is is argued that the empty set cannot truly be a member of anything c) true d) false e) true d)
I am more interested in questions than answers; dialog than dictation. If we can reasonably believe that there is not just a breach, but a fundamentally unclosable gap between the individual mind and the ultimate nature of the reality; the primordial thing in itself, then 'true' mystery does exist.

rigelrover
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Posted 10/02/09 - 11:05 AM:	#3
You should attempt to provide you own answers in the future, and let others correct you so you can learn from mistakes.
I am more interested in questions than answers; dialog than dictation. If we can reasonably believe that there is not just a breach, but a fundamentally unclosable gap between the individual mind and the ultimate nature of the reality; the primordial thing in itself, then 'true' mystery does exist.

Timothy
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Posted 10/02/09 - 11:35 AM:	#4
There's a 'trivial' proof showing that the empty set is a subset of every other set. X is a subset of Y if, and only if, all members of X are members of Y. Assume the empty set is not a subset of a given set Y. Then there is something that is a member of the empty set and not a member of Y. But there cannot be such a thing, since the empty set has no members, by definition. Therefore, the empty set is a subset of Y. Since Y was an arbitrary set, then this holds for all sets. rigelrover wrote: a) false b) true (but is is argued that the empty set cannot truly be a member of anything c) true d) false e) true Why is a) false and b) true? Let's call the empty set "q" and the membership operator "e". You're saying that q e {q} is false, but that q e {0, {q}} is true. It's backwards, a) is true and b) is false, since in b) q is not in {0, {q}}, because it's not the same thing to say q and {q}. That's why also c) is false, because 0 is not a member of { {0} }, and hence {0} is not a subset of it. And for the same reasons e) is also false rigelrover wrote: You should attempt to provide you own answers in the future, and let others correct you so you can learn from mistakes. Agreed.
"Neither Aristotelian nor Russellian rules give the exact logic of any expression of ordinary language; for ordinary language has no exact logic." P.F. Strawson

rigelrover
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Posted 10/02/09 - 11:44 AM:	#5
You are right Tim. Not sure why I did not think of that. Wish I had taken a class sometimes instead of just read the books.
I am more interested in questions than answers; dialog than dictation. If we can reasonably believe that there is not just a breach, but a fundamentally unclosable gap between the individual mind and the ultimate nature of the reality; the primordial thing in itself, then 'true' mystery does exist.

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