I know that the Mandelbrot set is connected, but what about its interior? It doesn't seem intuitively like it should be, but I can't find any information online confirming this. I can think of an intuitive argument for why the interior shouldn't be path-connected (moving for example from the main cardioid to one of the bulbs, a bifurcation must occur, so any path must cross the boundary) but a) this is completely unrigorous (it's an open problem whether every internal point is in a hyperbolic component, and this argument only works for hyperbolic components – what happens if there are infinitely many non-hyperbolic components?) and b) not being path-connected doesn't imply it's not connected. Can someone point me to some sources which discuss this at all?
Fractals – Is the Interior of the Mandelbrot Set Connected?
complex-dynamicsconnectednessfractalsreference-request
Related Solutions
Assume you have a continuous $g$ as stated by you.
Show that, for each $t \in [0,1]$, there exists a continuous function $\theta_{t} : [0,1]\rightarrow\mathbb{R}$ such that $\theta_{t}(0)=-\pi/2$ and such that $g(s,t)=|g(s,t)|(\cos\theta_{t}(s),\sin\theta_{t}(s))$. (Connectedness of $[0,1]$ can be helpful.) Then show that $\theta_{t}$ is unique by connectedness of $[0,1]$. Use known, simple contours for $f_{0}$ and $f_{1}$ in order to arrange for $\theta_{0}(1)=\pi/2$ and $\theta_{1}(1)=-3\pi/2$. Show that, for fixed $s \in [0,1]$, $\theta_{t}(s)$ is a continuous function of $t$ (because of homotopy,) and note that $\theta_{t}(1)$ has possible values $\pi/2\pm 2n\pi$ for $n=0,1,2,3,\ldots$. Use connectedness of $[0,1]$ to reach a contradiction.
Here's something I tried, thanks to Adam's comment for the basic idea for $p \ge 3$. This answer is missing some details, comments suggesting improvements are welcome, as would be other answers that fill in the gaps. It also relies on the fact (?) that:
$$\forall 0 \neq a \in \mathbb{C}, 0 \neq b \in \mathbb{C}, 1 < m \in \mathbb{N} . \exists x . |a e^{i x} + b e^{i m x}| \neq 1$$
Let $F(z, c) = z^2 + c$ with $F^{p+1}(z, c) = F^p(F(z, c), c)$. Now the boundary of a hyperbolic component can be parameterized by $\theta \in \mathbb{R}$ by the solution of the equation system: $$ F^p(z,c) = z \\ \frac{\partial}{\partial z}F^p(z,c) = e^{i \theta} $$ Now the question reduces to showing $c$ is of the form $c = c_0 + r_0 e^{i \phi}$ where $c_0 \in \mathbb{C}$ and $r_0 \in \mathbb{R}$ are constants and $\phi \in \mathbb{R}$.
$F^p(z,c) = z$ defines a polynomial of even degree $P(z) = 0$, whose constant coefficient is the product of its roots and is a polynomial in $c$ of degree $2^{p-1}$. The roots include those of $F^q(z, c) = z$ where $q | p$. Also, $\frac{\partial}{\partial z}F^p(z, c) = 2^p \Pi z_k$ where the $z_k$ are the $p$ roots in the periodic orbit of the desired solution $z$ (all $z_k$ are roots of $F^p(z, c) = z$, the remaining roots have lower period).
Case $p = 1$: $$ z^2 + c_0 + r_0 e^{i \phi} = z \\ \therefore z = \frac{1 \pm \sqrt{1 - 4(c_0 + r_0 e^{i \phi})}}{2} \\ \frac{\partial}{\partial z} = 2 z = e^{i \theta} $$
Now $|e^{i \theta}| = 1$ but $\exists x . |2 z| = |1 \pm \sqrt{x}| \neq 1$, so conclude that period $1$ component is not a perfect circle.
Case $p = 2$:
The equations reduce to $$ 4(1 + c_0 + r_0 e^{i \phi}) = e^{i \theta} $$ with obvious solution $c_0 = -1, r_0 = \frac{1}{4}, \phi = \theta$, so conclude that the period 2 component is a perfect circle.
Case $p = 3$:
The equations reduce to $$ 8 (c^3 + 2c^2 + c + 1) = e^{i \theta} \text{ where } c = c_0 + r_0 e^{i \phi}$$ For this to hold, the coefficients of $e^{i k \phi}$ must be zero for all $k > 1$. But setting $k = 3$ implies $r_0^3 = 0$ but we know that $r_0 > 0$ as hyperbolic components have non-empty interior. Contradiction, conclude that no period 3 component is a perfect circle.
Case $p > 3$:
Similarly to the $p = 3$ case, get a polynomial of degree $m > 1$ in $e^{i \phi}$ whose highest term has coefficient $r_0^m$. It remains to show that the polynomial really does have degree greater than $1$. The constant coefficient (product of roots) is a polynomial of degree $2^{p-1}$ in $c$, divided by the corresponding constant coefficient of all smaller divisors of the period gives:
$$m = 2^{p-1} - \sum_{q | p, q < p} 2^{q-1}$$
which solved numerically gives:
$$\begin{aligned} p & & & m \\ 1 & & & 1 \\ 2 & & & 1 \\ 3 & & & 3 \\ 4 & & & 5 \\ 5 & & & 15 \\ 6 & & & 25 \\ \vdots \end{aligned}$$
Finally, $m > 1$ for all $p \ge 3$ because $\exists q > 1 . q \nmid p$.
Best Answer
The interior of the Mandelbrot set is definitely not connected. It's rational parameter rays are known to lie in its complement and land at articulation points. This is described in great detail in Rational parameter rays of the Mandelbrot set by Dierk Schleicher. This image, taken from that paper, illustrates some of those rays: