Confusion About a Counterexample to a Statement

Question

I am currently facing a bit of confusion regarding a counterexample I provided for a statement in my mathematics class. My professor assigned us the task of proving the following statement false by providing a counterexample:

Statement: For all sets $S: S \not\subseteq \mathcal{P}(S)$

Here's the counterexample I came up with:

Counterexample: Let $S = \{1, 2\}$. In this case, the power set of $S$, denoted as $\mathcal{P}(S)$, is $\{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$.

Now, if we look at $S$ and $\mathcal{P}(S)$, we can see that $S = \{1, 2\}$ is indeed a subset of its power set, i.e., $S \subseteq \mathcal{P}(S)$.

However, my professor provided the following remark: $\{1,2\}$ is an element of $\mathcal{P}(S)$ but not a subset, keep looking."

I'm puzzled by this remark because I thought that every set is a subset of its own power set by definition. Could you please help me understand whether my counterexample is incorrect, or if there's something more subtle that I'm missing here? Is my professor mistaken in this case?

I appreciate any clarification or insights you can provide. Thank you

Answer 1

Subsets and elements are two different things. $S$ is a subset of itself, which means that $S$ is an element of its own power set. Maybe your professor's remark will be clearer once you understand that distinction.

But there is one set which is a subset of all sets, including the power set of some other set. Can you guess it?

Answer 2

I think we are making a big mistake here. If some set $S$ is a subset of another set $P$ then the set $S$ should have at least one element of $P$ in it. In your example it is seen that $P$ has an element $\{1,2\}$ in it as an element. If you look closely there is no element in the set $P$ that is also present in the set $S$. The confusion arose from the fact that $\{1\}$ is not same as the element $1$. When an element is covered by a set of curly bracers it becomes a set and not an element.