Topology, true or false? 
a) $\forall j : R^2 \to R^2$ $C^1$ function, $\exists k : R^2 \to R$ : $\nabla k(x) = j(x)$
b) $R^n$ has only one open vector subspace
c) $R^n$ has several compact vector subspace
d) Let $h : R^2 \to R$ a continuous function and $A$ a compact of $R$, $h^{-1} (A)$ is a compact

I did not manage to answer these questions satisfactorily
for b) i said "true" because if there is an open vector subspace, this open vector subspace is $R^n$
c) A compact is a bounded closed space, i said "true" but i'm not sure, i said that there are several singleton
d) I said "true" because $h^{-1} : A \to R^2$  a continuous function so $h^{-1}(A)$ is a compact
I don't know how to answer for a)
Thank you
 A: For b), subspaces are closed, and not open, except $\Bbb R^n$ considered as a subspace of itself, which is both open and closed. Kevin's answer, now deleted, had an argument with hyperplanes, another almost equivalent one: subspaces are defined by linear equations. The preimage of $\{(0,0,\dots,0)\}$ by a continuous function is closed.
For c), all subspaces are unbounded, except the null subspace.
For d), use my comment for b) to find a counterexample: the preimage is closed, but not necessarily compact ($h^{−1}$ is not a function!)

For a), the curl of the gradient is always zero. So it's not always possible to find a potential $k$ for a given vector field $j$. However, if $\mathrm{curl}\; j=0$ (i.e., the vector field is conservative), then such a $k$ exists. See Why curl free field implies existence of potential function? In two dimensions you have to consider the scalar curl.
To find a counterexample, it's thus enough to find a vector field $j(x,y)$ such that $\dfrac{\partial j_1}{\partial y}\ne \dfrac{\partial j_2}{\partial x}$.
Let $j$ be the linear map:
$$j(x,y)=\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}\cdot \begin{pmatrix}
x \\
y
\end{pmatrix}$$
That is, $j_1(x,y)=y$ and $j_2(x,y)=-x$.
Then, a potential $k$ for this vector field must satisfy:
$$\begin{eqnarray}
\dfrac{\partial k}{\partial x}&=&j_1(x,y)&=&y\\
\dfrac{\partial k}{\partial y}&=&j_2(x,y)&=&-x
\end{eqnarray}$$
From the first equation, you get that $k(x,y)=xy+a(y)$, for some function $a$ of only $y$. But then
$$\dfrac{\partial k}{\partial y}=x+a'(y)$$
And since $a'(y)$ does not depend on $x$, there is no way this can be equal to $-x$ for all $x$.
