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#1 2009-03-10 10:14:37

Onyx
Member
Registered: 2009-02-24
Posts: 48

kuratowskis theorem (a,b)={{a},{a,b}}

I have this question:

Let (a,b) denote {{a},{a,b}} then which of the following are true:

a.)

TRUE

b.)

FALSE

c.)

TRUE

d.)

TRUE

e.)

FALSE

f.)

FALSE

I'm not sure I understand the theorem, but I think my answers are right. Can someone check?

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#2 2009-03-10 10:51:51

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: kuratowskis theorem (a,b)={{a},{a,b}}

You goit (c) and (e) correct. Your other answers are incorrect, unfortunately.

(a) is false
(b) is true
(d) is false
(f) is true.

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#3 2009-03-10 11:12:30

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: kuratowskis theorem (a,b)={{a},{a,b}}

Wait a minute. Your answer to (c) is incorrect as well.

So (c) is false.

My, that was tricky. Even I got fooled. tongue

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#4 2009-03-10 23:58:13

TheDude
Member
Registered: 2007-10-23
Posts: 361

Re: kuratowskis theorem (a,b)={{a},{a,b}}

You're not using their definition.  Whenever you see numbers in parentheses you need to mentally replace them by the set.  For example, the first question asks if

.  To answer the question you need to replace (a,b) with {{a},{a,b}}.  You can see that the answer is actually false now (remember like Jane said that
).


Wrap it in bacon

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#5 2009-03-11 13:38:34

Onyx
Member
Registered: 2009-02-24
Posts: 48

Re: kuratowskis theorem (a,b)={{a},{a,b}}

Thanks for your replies, what you've said is initially what I thought, in which case I would have go them all, but then I changed my mind because of this reasoning:

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#6 2009-03-11 22:01:33

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: kuratowskis theorem (a,b)={{a},{a,b}}

Onyx wrote:


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#7 2009-03-11 22:06:40

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: kuratowskis theorem (a,b)={{a},{a,b}}

Onyx wrote:

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#8 2009-03-11 23:29:25

Onyx
Member
Registered: 2009-02-24
Posts: 48

Re: kuratowskis theorem (a,b)={{a},{a,b}}

Yes I know that, but since {a} is a set and {{a},{a,b}} is a set, and since a is an element of a set, I would have thought I used the relations correctly.

P.S. keep your hair on jane! smile

Last edited by Onyx (2009-03-11 23:30:37)

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#9 2009-03-12 00:02:41

TheDude
Member
Registered: 2007-10-23
Posts: 361

Re: kuratowskis theorem (a,b)={{a},{a,b}}

Onyx wrote:

Yes I know that, but since {a} is a set and {{a},{a,b}} is a set, and since a is an element of a set, I would have thought I used the relations correctly.

Remember that anything can be an element of a set, including other sets.  In the case of {{a}, {a, b}} every element of this set is another set.  Since a by itself is not a set it cannot be an element of {{a}, {a, b}}.


Wrap it in bacon

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#10 2009-03-12 04:41:53

Onyx
Member
Registered: 2009-02-24
Posts: 48

Re: kuratowskis theorem (a,b)={{a},{a,b}}

Ok I get it now. my reasoning doesn't work since

and
I see what you were saying.

Thanks

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#11 2009-03-12 23:17:40

Onyx
Member
Registered: 2009-02-24
Posts: 48

Re: kuratowskis theorem (a,b)={{a},{a,b}}

Hi, I've got another question now regarding kurwatowskis theorem. I need to prove:

It seems like it should be so obvious, but I'm having trouble writing a meaningful proof for it, without actually just replacing a with u and b with v.

I'm given the hint that I need to consider the two cases where a=b and

My thoughts are that these cases are:


since a=u and b=v,

I feel like I'm making a mess of it, please help!

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#12 2009-03-14 07:16:40

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

Re: kuratowskis theorem (a,b)={{a},{a,b}}

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#13 2009-03-15 10:16:05

Onyx
Member
Registered: 2009-02-24
Posts: 48

Re: kuratowskis theorem (a,b)={{a},{a,b}}

Thanks alot thats a great help.

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