whether the proof is true or false, $\partial\bar{\Omega}= \partial \Omega$, approximated by polynomials, simply-connected $\Omega\subset\Bbb C$ is a bounded domain, $\partial\bar{\Omega}= \partial \Omega$. For every $f(z)$ that is continuous on $\bar{\Omega}$ and analytic in $\Omega$ ,$f(z)$ can be approximated by polynomials, uniformly on $\bar{\Omega}$.
(1). Show that $\Omega$  is  simply-connected.
(2). IF the boundedness assumption on $\Omega$ is dropped, is  $ \Omega$   simply-connected?
(3). Find a bounded simply connected domain $\Omega$ that satisfies $\partial\bar{\Omega}= \partial \Omega$ and a $f(z)$ that is continuous on $\bar{\Omega}$ and analytic in $\Omega$ , But $f(z)$ can not  be approximated by polynomials, uniformly on $\bar{\Omega}$

$\Omega$  is  simply-connected if the winding numbers $n(\gamma,z_0)=0$ for every point  $z_0$ in $\Bbb C\setminus \Omega$ and every closed, piecewise smooth path $\gamma$ in $\Omega$.
I guess (2) is  correct.   whether the following  proof is true or false? Thanks a lot
If $z_0\in \Bbb C\setminus \bar{\Omega}$, then $f(z)=\dfrac{1}{z-z_0}$  is continuous on $\bar{\Omega}$ and analytic in $\Omega$ ,$f(z)$ can be approximated by polynomials  $p_n$, uniformly on $\bar{\Omega}$ so
$$n(\gamma,z_0)=\dfrac{1}{2\pi i}\int_{\gamma}\dfrac{1}{z-z_0}dz=\dfrac{1}{2\pi i}\lim_{n\to\infty} \int_{\gamma}p_n(z)dz=0 $$
If $z_0\in \partial \Omega$, because $\partial\bar{\Omega}= \partial \Omega$, so there exist ${z_n}\in\Bbb C\setminus \bar{\Omega}$, such that $\lim z_n=z_o$, with the consequence that $f_n =\dfrac{1}{z-z_n}$ converges normally to $f$ in  $ \Omega$. Since $z_n\in \Bbb C\setminus \bar{\Omega}$, so $\int_{\gamma}\dfrac{1}{z-z_n}dz=0$, then
$$n(\gamma,z_0)=\dfrac{1}{2\pi i}\int_{\gamma}\dfrac{1}{z-z_0}dz=\dfrac{1}{2\pi i}\lim_{n\to\infty} \int_{\gamma}f_n(z)dz=0 $$
so,  for every point $z_0$ in $\Bbb C\setminus \Omega$ and every closed, piecewise smooth path $\gamma$ in $\Omega$ , $n(\gamma,z_0)=0$, hence  $\Omega$  is  simply-connected
 A: Partial answer solving points 1,3 only (have to think more on pooint 2)- the condition $\partial\bar{\Omega}= \partial \Omega$ means that $C-\Omega$ is either connected or has a bounded open component $V$ and assuming by contradiction we are in the second case, for any $a$ inside such a bounded open component, 
$$
f(z)=\frac{1}{z-a}
$$ 
cannot be uniformly approximated by polynomials on $\Omega$ as otherwise taking 
$$
M=\max|a-z|,\quad z \in \bar \Omega,
$$ 
then $0<M<\infty$ and for any polynomial satisfying 
$$
|P(z)-f(z)| < \frac{1}{M} \le \frac{1}{|z-a|} \implies|(z-a)P(z)-1|<1,\quad z \in \bar \Omega.
$$
 But $\partial V$ is included in $\bar \Omega$ hence by maximum modulus said relation holds on $V$ and letting $z=a$ we get a contradiction $1<1$!
For point 3, take any moon-shaped domain (simply connected with boundary formed by two circles or analytic Jordan curves inner tangent at a point - for example the domain given by $|z|<2, |z-1|>1$) and the proof in point 1 still applies
Later edit- I think that any multiply connected domain with non-degenerate boundary components - this is the condition $\partial\bar{\Omega}= \partial \Omega$ - bounded or not must have a Jordan curve $J$ enclosing a boundary component and then the inside of the Jordan curve has points where the previous construction applies to show that polynomials cannot approximate uniformly analytic functions there (we need just the distance to $J$ to be finite and non zero, to apply the same method and get a contradiction) but I am not 100% certain as the complex plane has some subtle topological properties, so for point 2 I think the result remains correct with the caveat above
