I'm asked to state whether a statement is true or false and the following is one that is confusing to me: $$\{0\}\subseteq\mathbb{P}(\{0,1,2\})$$ I thought this would be true, but the book says this is false.
Wouldn't $\mathbb{P}(\{0,1,2\}) = \{\{0\},\{1\},\{2\},\{0,1\},\{0,2\},\{1,2\},\{0,1,2\},\emptyset\}$?
Is $\{0\}$ a subset of $\mathbb{P}(\{0,1,2\})$?
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elementary-set-theory
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4Your $\Bbb P$ is indeed correct. Your conclusion that $\{0\} \subseteq \Bbb P$ is incorrect. – 2017-02-15
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2Note the difference between $\in$ and $\subseteq$. – 2017-02-15
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0Note : {0,1,2} is also in P. Apart from I agree with kennytm, it's all about being element and being included. – 2017-02-15
1 Answers
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$A\subseteq B$ if and only if every element of $A$ is an element of $B$. This is easy to check since $\{0\}$ has only one element - $0$. You've almost correctly listed the powerset (you're missing $\{0,1,2\}$), and we notice that none of the listed elements are $0$. One of them is $\{0\}$, but that's different and is probably what's causing you confusion. It is true that $\{\{0\}\}\subseteq P(\{0,1,2\})$ because in that case the element is $\{0\}$.
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2So in other words, the sentence $\{0\} \in P(\{0,1,2\})$ is true, but $\{0\} \subseteq P(\{0,1,2\})$ is **not**. – 2017-02-15