----- Original Message -----
From: Bruno Marchal
To: [email protected]
Sent: Tuesday, July 07, 2009 2:07 PM
Subject: Re: The seven step series
On 07 Jul 2009, at 16:18, m.a. wrote:
Bruno,
I'm not entirely sure of these answers, but I think I learn more from your
corrections than from pondering the rules to the point where I confuse myself.
m.a.
Do you remember, I asked you to give me all the subsets of {1, 2}. That is,
all the sets which are included in {1, 2}. You gave me the correct answer:
those subsets are { }, {1}, {2}, {1, 2}. You see that the set {1, 2} has 2
elements, and 4 subsets. But then I asked to give me the set of all subsets of
{1, 2}.
{1, 2} has four subsets, and it is natural to make that many a one, by
considering *the* set of all subsets of {1, 2}. The answer is:
{{ }, {1}, {2}, {1, 2}}
Considering all subsets of a set is a rather important operation, which we
will meet more than one times in the sequel. Given its importance
mathematicians gave it a name. It is the power operation. Later I will be able
to explain why it is called power.
It is an UNARY operation, which means it applies on ONE set. (Intersection,
and union are BINARY operations, they need two sets to work on).
So (power x) = {y such-that y is included in x}, by definition.
For example:
(power {1, 2}) = {{ }, {1}, {2}, {1, 2}}
Here are the three promised exercises. Compute
(power {1}) = ? {{ }, {1}}
(power {1, 2, 3}) = ? {{ }, {1}, {2}, {3}}
(power { }) = ? {{ }}
Take your time,
Bruno
http://iridia.ulb.ac.be/~marchal/
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