# Power set, belongs or inclusion?

My doubt is relative to the power set and the notation.

If we have a set $$A = \{a,b,c\}$$, its power set is $$\mathcal{P}(A) = \{\emptyset,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},A\}$$

So we get rightly these notations

$$a \in A$$, $$\{a\} \subset \mathcal{P}(A)$$, $$\{a\} \in \mathcal{P}(A)$$?

• Why $\{a\}\subset P(A)$? Is $a$ a member of $P(A)$? – almagest Sep 27 '19 at 10:33

You already wrote down correctly what $$\mathcal{P}(A)$$ is. So you can just check manually.

• Definitely $$a \in A$$, because that is how you defined $$A$$.
• For $$\{a\} \subset \mathcal{P}(A)$$ to hold, we must have that every element of $$\{a\}$$ is an element of $$\mathcal{P}(A)$$. So that would mean that $$a \in \mathcal{P}(A)$$, but as you can see $$a$$ never appears as an element of $$\mathcal{P}(A)$$. So $$\{a\} \not \subset \mathcal{P}(A)$$.
• Finally, $$\{a\}$$ does appear as an element in $$\mathcal{P}(A)$$, so indeed $$\{a\} \in \mathcal{P}(A)$$.

When getting confused about notation, it is often best to just write out exactly what the notation means according to its definition.

• Therefore is it right $\{\{a\},\{b\}\} \subset \mathcal{P}(A)$? – LH8 Sep 27 '19 at 10:41
• @LH8 Yes! I think you got it now. – Mark Kamsma Sep 27 '19 at 10:41
• However, $\left{\left{a\right}right} \subset \mathcal{A}$ – Samuel Bodansky Sep 27 '19 at 10:46
• And also it is true $\{\{a\}\} \subset \mathcal{P}(A)$? – LH8 Sep 27 '19 at 10:51
• @LH8 Yes, that is actually what Samuel tried to write in the comment above (something went wrong with formatting). – Mark Kamsma Sep 27 '19 at 10:53

We have $$a\in A$$ and also $$\{a\}\in\mathcal P(A)$$.

Further we will have $$\{a\}\subseteq\mathcal P(A)$$ if and only if $$a\in\mathcal P(A)$$.

Looking at $$\mathcal P(A)$$ we conclude that this occurs if and only if at least one of the following conditions is satisfied:

• $$a=\varnothing$$
• $$a=\{b\}$$
• $$a=\{c\}$$
• $$a=\{b,c\}$$

Accepting the axiom of regularity we excluded on forehand that the conditions $$a=\{a\},a=\{a,b\},a=\{a,c\}$$ and $$a=A=\{a,b,c\}$$ are satisfied.

This because this axiom "forbids" that $$a\in a$$.

Note that this concerns special cases. In general if $$A=\{a,b,c\}$$ then it is not true that $$\{a\}\subseteq\mathcal P(A)$$. The answers of José and Mark are in that context.

Yes, $$a\in A$$, and, yes, $$\{a\}\in\mathcal P(A)$$. But it is not true that $$\{a\}\subset\mathcal P(A)$$.