# Re: The seven step series

From: m.a. <marty684.domain.name.hidden>
Date: Tue, 7 Jul 2009 20:28:17 -0400

----- Original Message -----
From: Bruno Marchal
To: everything-list.domain.name.hidden
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 02 elements, and 04 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 { }) = ? {{ }}

Bruno

http://iridia.ulb.ac.be/~marchal/

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Received on Tue Jul 07 2009 - 20:28:17 PDT

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