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Something I'm failing to understand from Halmos "Naive Set theory" book.

If $\Phi $ is a collection of subsets of a set E (that is, $\Phi$ is a subcollection of $\rho (E)$), then write

First of all I would like to point out that letters Phi and rho aren't the ones used in Halmos book. I do not recognize the letters from the book so I had no option but to pick my own.

$$\rho(E) = \{X:X\subset E\}$$

This is the definition of $\rho (E)$ from the book.

My question is: What is the difference between $\rho (E) $ and $\Phi$?

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    $\begingroup$ (The '$\Phi$' used in the text is a calligraphic 'C', i.e. $\mathcal C$, I think.) $\endgroup$
    – Watson
    Jan 23, 2016 at 17:40
  • $\begingroup$ What you're writing as $\rho$ is $\mathcal{P}$, which is \mathcal{P}. $\endgroup$
    – BrianO
    Jan 23, 2016 at 19:04

1 Answer 1

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The set $\rho(E) = \mathcal P (E)$ is the set of all the subsets of $E$, while $\Phi$ is a collection of subsets of $E$ (which may not contain all of them). In other words : $\Phi \subseteq \rho(E)$.

For instance, if $E =\{1,2\}$, then the power set of $E$ is $\rho(E) = \{\varnothing, \{1\}, \{2\}, E\}$ and you can take $\Phi = \{\{2\}, E\}$ as a collection of subsets of $E$ (which is not the whole power set).

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    $\begingroup$ If $\Phi$ was an empty set would it still be a collection of subsets of E? Or does it have to be plural? $\endgroup$ Jan 23, 2016 at 17:08
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    $\begingroup$ If $\Phi = \varnothing$, then it is an (empty) collection of subsets of $E$, since $\Phi = \varnothing \subseteq \mathcal P (E)$. $\endgroup$
    – Watson
    Jan 23, 2016 at 17:10
  • $\begingroup$ For any set $Z$ whatsoever, the empty set is a subset of $Z$. So in particular, if $\Phi$ is the empty set, then $\Phi$ is a subset of of the set $\rho(E)$. When using the English phrase "a collection of subsets of $E$", you should not focus too much on the use of the plural; it would be too tedious to have to say "a collection of subsets of $E$ that may include just one subset or maybe none at all" every time you wish to use this phrase. $\endgroup$
    – Lee Mosher
    Jan 23, 2016 at 17:12

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