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I am now struggling with the basics of Formal Semantics, which is hard to understand without knowing something about the Naive Set Theory. Please help me out.

The task is to explicitly write down the set by enumerating its members:

{a} ∩ P({a})

I incorrectly thought that the answer is {a}, but it turned out to be Ø. I don't know what to make of it. The Naive Set Theory by Halmos does not provide me with the clear answer, but it is much more probable that I can't find one.

Please help.

Respectfully, Z

Zimbardo
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4 Answers4

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The other posted answers have ignored the possibility that $a=\emptyset.$ (Added: @celtschk mentioned this possibility and posted it while I was writing my answer.)

The members of $\mathscr{P}(\{a\})$ are precisely $\emptyset$ and $\{a\}.$

The only member of $\{a\}$ is $a.$

So is $a$ (the only member of $\{a\}),$ a member of $\mathscr{P}(\{a\})?$ We need to check the two members of $\mathscr{P}(\{a\})$ to see if $a$ could be either of them: it's true that $a$ doesn't equal $\{a\},$ but it's possible that $a$ equals $\emptyset.$

As a result,

$$\{a\}\cap\mathscr{P}(\{a\})=\begin{cases}\{\emptyset\},&\text{if }a=\emptyset,\\\emptyset,&\text{if }a\ne\emptyset.\end{cases}$$

Mitchell Spector
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$\{a\}\cap \mathcal P(\{a\})$ contains everything that is both element and subset of $\{a\}$. The only element of $\{a\}$ is $a$, and the subsets of $\{a\}$, and thus the elements of $\mathcal P(\{a\})$, are $\emptyset$ and $\{a\}$.

Now it is possible that $a=\emptyset$, and in that case, your solution would be right. However obviously it is (either explicitly or implicitly) assumed that $a\ne\emptyset$. Since also $a\ne\{a\}$ (at least using standard set theory), clearly $\{a\}$ and $\mathcal P(\{a\})$ don't have a common element, and thus the intersection is empty.

celtschk
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You start from the definition of the power set, which is a set that includes all the subsets of its argument. $\{a\}$ has two subsets, $\emptyset$ and $\{a\}$, so $P(\{a\})=\{\emptyset, \{a\}\}$. Note that $a$ is not an element of this powerset, so $\{a\}\} \cap \{\emptyset, \{a\}\}=\emptyset$

Ross Millikan
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Explicitly writing out $\mathcal{P}(\{a\})$ (the set of subsets of $\{a\}$) we have $\mathcal{P}(\{a\}) = \{\emptyset, \{a\}\}$

Comparing the sets $\{a\}$ and $\{\emptyset, \{a\}\}$ we see that these have no elements in common. On the left, it has the element $a$. On the right it has the elements $\emptyset$ and $\{a\}$. Note that the element $\{a\}$ is not the same thing as the element $a$.

As such, $\{a\}\cap \{\emptyset,\{a\}\}=\emptyset$

JMoravitz
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