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I'm trying to self-learn math and I'm starting with naïve set theory. The only problem is that the book I'm using lacks a solution textbook and I don't know if the solution I gave to this problem is right or not.

The problem is this and in parenthesis I put my answer: "If $A =\{0,3,5\}$ and $\mathcal{P}(A)$ is the power set of $A$, determine which of those statements are true:

  • $0 \in A$ (True); 0 is an element of the set A and so it's a member of it
  • $\{5\} \in A$ (False); $\{5\}$ is a subset and not a member of $A$
  • $\varnothing \in A$ (false);$ \varnothing$ is the empty set and not a member of the set $A$
  • $\varnothing \subseteq A$ (False);$ \varnothing$ is a subset of every set(or i'm confusing it with the power set?)
  • $\varnothing \subset A$ (false);Same as up
  • $\varnothing \notin A$ (true); $ \varnothing$ is not a member of $A$
  • $0 \in \varnothing$ (false); $ \varnothing$ is the empty set and doesn't have elements
  • $\varnothing \in \{ \varnothing\}$ (true); $ \varnothing$ is an element of the set
  • $\varnothing \subseteq \{ \varnothing\}$ (false);$ \varnothing$ is not a subset(or since,there is only one element it can be considered a subset?)
  • $\varnothing \in \mathcal{P}(A)$ (true); $ \varnothing$ is an improper set and an element of every power set
  • $\varnothing \subseteq \mathcal{P}(A)$ (false); $ \varnothing$ is not a subset since the power set contains only elements
  • $\{ \varnothing\} \subseteq \mathcal{P}(A)$ (false); There isn't a subset that have as only element the empty set in the power set
  • $\{ \varnothing\} \in \mathcal{P}(A)$ (false)."

I added the reasoning behind my choice.

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  • $\begingroup$ See meta.math.stackexchange.com/questions/5020/… for some formatting guidelines; as it stands this is very hard to read. Also, for each of your answers, please explain your reasoning - this will help people to give much better suggestions. $\endgroup$ – Nate Eldredge Jan 19 '15 at 19:38
  • $\begingroup$ @mrp I read it,but it seems i made everything wrong.I'm going to add the reasoning,and thank you for the edit and the help. $\endgroup$ – Mik89 Jan 19 '15 at 19:42
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    $\begingroup$ That's not quite a word problem. $\endgroup$ – Asaf Karagila Jan 19 '15 at 19:46
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You have a few mistakes:

$\emptyset \subseteq A$ is $\textbf{true}$ as every single element of $\emptyset$ (there are none) are included in $A$.

$\emptyset \subset A$ is $\textbf{true}$ for the same reasoning as above and because $\emptyset \neq A$

$\emptyset \subseteq \{\emptyset\}$ is $\textbf{true}$ for above reasons

$\emptyset \subseteq \mathcal{P}(A)$ for above reasons

$\{\emptyset\} \subseteq \mathcal{P}(A)$ is $\textbf{true}$ because every element of $\{\emptyset\}$ is contained in $\mathcal{P}(A)$ as $\emptyset \in \mathcal{P}(A)$.

Remember that the power set is a set of sets, i.e if $C = \{a,b,c\}$ then

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

It seems like you have misunderstood the subset relation $\subseteq$. We write $A \subseteq B$ whenever every element of $A$ is included in $B$. If $A$ has no elements, this is trivially true.

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  • $\begingroup$ Thank you for the answer.And you are right about the subset relation related to my confusion. $\endgroup$ – Mik89 Jan 21 '15 at 12:50

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