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Let's assume that we have a set:

$$ X = \{a, b, c\} $$ Is it true, that a Boolean algebra of this set is like below?

$$ P(X) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\} $$

If yes, what are subalgebras of this Boolean algebra? I need to write all of them.

JohnDoe
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    Not quite. You should just have $\emptyset$ where you have ${\emptyset}$. There is a difference between $\emptyset$ and ${\emptyset}$. Also, this is usually called the power set of the set ${a,b,c}$, although it is in fact a Boolean algebra with the operations of union, intersection, and complementation. ${}\qquad{}$ – Michael Hardy Jan 17 '15 at 17:55
  • Thanks for a comment. I know this, but I made a mistake due to copy/paste braces. What with subalgebras then? – JohnDoe Jan 17 '15 at 17:56

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The empty set itself is always part of a subalgebra, and all by itself it makes one.

If you additionally include any one of the other seven elements, you have a subalgebra of order two.

Now imagine including two elements beyond the empty set, $A$ and $B$. This order-three set can't be a subalgebra because either $A\cup B$ is new, or wlog $A\subset B$, in which case $B\setminus A$ is new. So consider the $\binom{7}{2}=21$ pairs of non-empty elements you could include alongside the empty set, and fully list out what subalgebra they generate with $\cup$, $\cap$, and $\setminus$. There will be at most seven such subalgebras since they end up being at least size 4 and there fore each have three such pairs.

Lastly, can you have a subalgebra of order five, six, or seven? Imagine you have such a subalgebra. After accounting for the empty set, consider two more elements from this supposed subalgebra. These alone will generate one of the subalgebras listed from the previous step. For each of those, see what happens if you even include one more element and take the closure with respect to $\cup$, $\cap$, and $\setminus$.

2'5 9'2
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