If a set is open (closed) in $R^m$ is it also open (closed) in $R^{m\pm 1}$? I am trying to prove the following two propositions true or false:
1 - If A is an open (closed) subset of $\mathbb{R}^{m+1}$, then 
\begin{equation}
\{x ∈\mathbb{R}^m : (0,x) ∈ A\} 
\end{equation}
is open (closed) in $\mathbb{R}^m$.
2 - If A is an open (closed) subset of $\mathbb{R}^{m}$, then 
\begin{equation}
\{(0,x) ∈\mathbb{R}^{m+1} : x ∈ A\} 
\end{equation}
is open (closed) in $\mathbb{R}^{m+1}$.
To prove the first proposition true I have proceed in the following way:
Since A is open $\forall(0,y)\in A$, $\exists r>0$ such that $B((0,y),r) \subset A$. Now, every $(0,z)\in B((0,y),r)$ also belongs to $A$ and:
\begin{equation}
 d((0,z),(0,y))<r
\end{equation}
\begin{equation}
 \sqrt{(0-0)^2+(z_1-y_1)^2+...+(z_m-y_m)^2}<r
\end{equation}
\begin{equation}
 \sqrt{(z_1-y_1)^2+...+(z_m-y_m)^2}<r
\end{equation}
\begin{equation}
 d(z,y)<r
\end{equation}
Hence we can construct the ball $B(y,r) \subset \mathbb{R}^m$. Since $(0,z) \in A$, then $z \in \{x ∈\mathbb{R}^m : (0,x) ∈ A\} $, and consequently all the points in the ball B(y,r) also belong to the set, this is $B(y,r) \subset \{x ∈\mathbb{R}^m : (0,x) ∈ A\}$. With this it is proven that the set $\{x ∈\mathbb{R}^m : (0,x) ∈ A\}$ is open.
By the same reasoning and starting from the fact that $A^c$ is open I have also proven it true for the case in which A is close.
Could you confirm me if these proofs are correct?
For the second proposition in the case in which A is open I have proven it false directly by a counterexample. However, I am a bit lost for the case in which A is closed. I have been analyzing some examples and I have the intuition that it is true but I am not able to formally prove it. Could you give me some hints about how to proceed?
Thanks for your help!
 A: 2).
If $A = (a,b)$ is open then $B = \{(0, x)|x \in (a,b)\}:= \{0\}\times (a,b)$ is clearly not open in $\mathbb R^2$ as every open neighborhood of $(0,x)$ will contain some $(\epsilon, x')$ where $\epsilon \ne 0$ so  $(\epsilon, x') \ne \{0\}\times (a,b)$ so $(0,x) \in \{0\}\times (a,b)$ is not an interior point.
So 2) is not true for open sets.
....
But if $X $ is closed in $\mathbb R^m$ then $W= \{0\}\times X$ is closed in $\mathbb R^{m+1}$.
Let $(w,z) \in \mathbb R^{m+1}; i.e. w \in \mathbb R; z \in \mathbb R^m$ where $w \ne 0$.  Then let $0 < \epsilon < |w|/2$.  If $N((w,z), \epsilon) = \{(a,b)\in \mathbb R^{m+1}| d((a,b),(w,z)) < \epsilon\}$ then if $(a,b) \in N((w,z), \epsilon)$ the $|a - w| = d((a,z),(w,z)) \le d((a,b),(w,z)) < \epsilon < |w|/2 < |w| = |0 - w|$ so $a\ne 0$ so $(a,b) \not \in \{0\}\times X$.  
So $(w,z)$ is not a limit point of $W$.  
If $z \in \mathbb R^m$ is not a limit point of $X$ then there exists an $\epsilon > 0$ so that $N(z,\epsilon)$ contains no point of $X$.  Then $N((0,z),\epsilon) \subset \mathbb R \times N(z,\epsilon)$ contains no point of $W \subset \mathbb R \times X$.
So $(0, z)$ is not a limit point of $W$.
So the only limit point of $W$ must be of the form $(0, z)$ where $z$ is a limit point of $X$.  As $X$ is closed, $z \in X$ so $(0,z) \in W$.  So all limit points of $W$ are in $W$.  
So $W$ is closed.
So 2) is true for closed sets.
