I don't have complex analysis at my beck and call, and I only have a low level of knowledge in topology, but I need to prove that this metric space (for any real $r$ and $R$ with $r < R$)$$ X = \{ (x, y) \in \mathbb{R}^2 \ | \ r \leq x^2 + y^2 \leq R \}$$
with the Manhattan metric $d((x_1, y_1), (x_2, y_2)) = |x_1-x_2| + |y_1-y_2|$ is not simply connected.
I've already prooven that it's path connected, and now I need to show there are some points $P$ and $Q$ with two paths between them such that one cannot be continuously 'morphed' into the other.
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What I have so far is as follows:
I take $P = (0, r)$ and $Q = (0, -r)$, with $f_0$ being a path from $P$ to $Q$ going clockwise around the circle radius $r$ and $f_1$ being much the same but going counterclockwise.
Now I assume there is a function $g : [0, 1]^2 \to X$ such that:
- $g(s, 0) = f_0 (s)$
- $g(s, 1) = f_1 (s)$
- $g(0, t) = P$
- $g(1, t) = Q$
To get the final result, I need to show that this function cannot be continuous, but for the life of me I cannot.
For some context, these are the topics which have been visited during the course, roughly in order of recentness.
- Multiple connectedness
- Simple connectedness
- Pathwise connectedness
- Interior points
- Boundary points
- Open sets
- Compactness
- Complete metric spaces
- Bounded metric spaces
- Totally bounded metric spaces
- Closed sets
- Closure of a metric space
- Limit points
- Cauchy sequences
- Convergence
- Continuity
- Metric space quivalence
- Metric equivalence
Edit: I might have an argument that works, though it's far from rigourous. We can shrink $R$ to be as close to $r$ as we want, so we can essentially constrain the annulus down to a circle and thus force any path from the left side of the circle to the right side to go through $P$ or $Q$.
So holding $s \in (0, 1)$ constant and varying $t$ must produce a path through $P$ or $Q$ for any $s$. If for some $s$ it passes through $P$ and for some other $s$ it passes through $Q$, then there must be $s_0$ such that $\forall \epsilon >0 \ \exists \delta \leq \epsilon$ st. $s_0$ produces a path through $P$ and $s_0 + \delta$ produces a path through $Q$.
Now we can consider the path $g(s, \frac{1}{2})$, and note that it must have a discontinuity at $s_0$.
Now consider the case where all $s$ produce paths through only one of $P$ or $Q$. WOLOG: $P$. Now, at $t = \frac{1}{2}$, $s$ arbitrarily close to $1$ are mapped away from $Q$, but $1$ is always mapped to $Q$ by definition, so $g(s, \frac{1}{2})$ has a discontinuity at $s=1$. Therefore $g(s, t)$ is not continuous.
This argument is definitely iffy to me, if no one has their own argument (using sufficiently low level concepts), then criticism on the above would be appreciated.
Best Answer
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.