Q:

Subset, Not a subset, And Elements

Question $1$) Write $\subseteq$ or $\not\subset$:
$\Bbb N\underline{}\Bbb Q$
$\Bbb N\underline{}\wp(\Bbb R)$
$\varnothing\underline{}\Bbb Z$
$\sqrt2\underline{}\Bbb R$
$\Bbb Z \cup [-1,1]\underline{}[-2,2]$
Question $2$) Write $\subseteq$ or $\in$:
$(3,5)\underline{}[3,5]$
$[-1,4]\underline{}(-1,4)$
$\{\varnothing\}\underline{}\wp(\Bbb Q)$
$\Bbb N \times\Bbb Z\underline{}\Bbb Z \times\Bbb N$  
I'm not sure if my answers are correct:
Question $1$: $\subseteq,\not\subset,\subseteq,\subseteq,\not\subset$
Question $2$: $\subseteq,\in,\subseteq,\subseteq$

A:

1a) Every Natural Number $N \in \mathbb{N}$ is also a rational: we can just write it as $\frac{N}{1}$. So $\mathbb{N} \subseteq \mathbb{Q}$.
1b) $\mathbb{N} \nsubseteq \wp\mathbb{R}$, because every element in $\wp\mathbb{R}$ is a set, and this would imply every $n \in \mathbb{N}$ is a set, which is not true. However, $\mathbb{N} \in \wp\mathbb{R}$.
1c) The empty set is a subset of any set. (You should prove or look up this fact.) So $\emptyset \subseteq \mathbb{Z}$.
1d) $\sqrt{2} \nsubseteq \mathbb{R}$. $\sqrt{2}$ is not a set. It is, however, a member of the set of the real numbers: $\sqrt{2} \in \mathbb{R}$.
1e) $\mathbb{Z} \cup [-1,1] \nsubseteq [-2,2]$. The union of $\mathbb{Z}$ and $[-1,1]$ obviously contains elements not contained by $[-2,2]$.
2a) $(3,5) \subseteq [3,5]$. You should read about open and closed intervals.
2b) $[−1,4]\_(−1,4)$ does not make sense given the options provided. $[−1,4]\nsubseteq(−1,4)$ because $[-1,4]$ contains $-1,4$ which are not contained by the open interval $(-1,4)$. However, the interval $[-1,4]$ can also obviously not be an element of the interval $(-1,4)$.
2c) $\{\emptyset\} \subseteq \wp\mathbb{Q}$. The set containing the empty set is always contained in any powerset. You should play with this statement to prove its truth.
2d) $\mathbb{N} \times \mathbb{Z} \nsubseteq \mathbb{Z} \times \mathbb{N}$. The definition of the cartesian product $A\times B = \{(a,b) \mid a \in A, b \in B\}$ should provide the necessary intuition. (Obviously, $\mathbb{N} \times \mathbb{Z} \notin \mathbb{Z} \times \mathbb{N}$ - so this statement also does not make sense given the options provided.)

