$\mathbb C^2_{0} = \{(a,b)\in X: a\ne 0, b\ne 0\}$ is an open path-connected subset of $\mathbb C^2$

Question

If $X$ is an open path-connected subset of $\mathbb C^2$, prove that $$\mathbb C^2_{0} = \{(a,b)\in X: a\ne 0, b\ne 0\}$$ is also an open path-connected subset of $\mathbb C^2$.

1. $\mathbb C^2_{0}$ is open.

$\mathbb C^2_{0}$ is open in the subspace topology on $X$, from which the claim follows.

2. $\mathbb C^2_{0}$ is path-connected.

Take $x,y\in \mathbb C^2_{0}$ (of course, $x$ and $y$ are tuples, and I am lazy). I need to find a continuous function $f: [0,1]\to \mathbb C^2_{0}$ such that $f(0) = x$ and $f(1) = y$. Since $X$ is path-connected, there exists $g: [0,1]\to X$ with $g(0) = x$ and $g(1) = y$. My sense is that we must modify $g$ in some way (restrict to $\mathbb C^2_{0}$, or something) to construct $f$, but I am stuck here.

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Possible Generalization: I wonder if the following generalization is true, for $n\in \mathbb N$:

If $X$ is an open path-connected subset of $\mathbb C^n$, is $$\mathbb C^n_{0} = \{(a_1,a_2,\ldots,a_n)\in X: a_i \ne 0, 1\le i\le n\}$$ also an open path-connected subset of $\mathbb C^n$?

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I was able to prove path-connectedness for a very special case, i.e. when $n=1$ and $X\subset \mathbb C$ is convex. I doubt if a similar idea would work for the more general cases ($X$ may not be convex, $n=2$ or greater, etc.) Take $n=1$, and let $X$ be a convex subset of $\mathbb C$. If $0\notin X$, then $C_0^1 = X$, so there is nothing to do. Assume $0\in X$. Then, $C_0^1 = X\setminus \{0\}$. Consider $x,y\in C_0^1$, and the straight line path in $X$ given by $f:[0,1]\to X$ such that $f(t) = (1-t)x + ty$ for $t\in [0,1]$. If $f(t)\ne 0$ for all $t\in [0,1]$, then $f:[0,1]\to C_0^1$ is the required path in $C_0^1$. If $f(t) = 0$ for some $t\in [0,1]$ (there is at most one such $t$), then consider $z\notin f([0,1])$. Define $g_1:[0,1]\to C_0^1$ by $g_1(t) = (1-t)x + tz$ and $g_2:[0,1]\to C_0^1$ as $g_2(t) = (1-t)z + ty$. Concatenate $g_1$ and $g_2$ to get $g:[0,1]\to C_0^1$. $$g(t) = \begin{cases}g_1(2t) & t\in [0,1/2]\\ g_2(2t-1) & t\in [1/2,1] \end{cases}$$ $g$ is the required path between $x$ and $y$ in $C_0^1$.

Thanks!

Answer 1

$\newcommand{\Cpx}{\mathbf{C}}$Here are some suggestions.

1. fish has hinted at the "right" proof of openness.

2. For path connectedness, the intuition is we're removing two complex lines from $\Cpx^{2}$. Since these have real codimension $2$, they don't separate $X$. To prove this, we might

• Show that every continuous path $f$ in the connected open set $X$ can be homotoped to a polygonal path in $X$. Hint:

One way to do this rigorously is to cover the image of $f$ with open polydisks $D$ contained in $X$. Use compactness to select a finite subcovering $(D_{k})_{k=0}^{n}$ in such a way that $f(0) \in D_{0}$, $f(1) \in D_{n}$, and $D_{k-1} \cap D_{k}$ is non-empty for $1 \leq k \leq n$; that is, construct a "chain of polydisks" in $\Cpx_{0}^{2}$ covering the image of $f$. Let $x_{0} = f(0)$, $x_{n+1} = f(1)$, and for each $k$ with $1 \leq k \leq n$, pick a point $x_{k}$ in $D_{k-1}\cap D_{k}$ and on the image of $f$, say $x_{k} = f(t_{k})$ for some $t_{k}$ in $[0, 1]$. Note that both $x_{k}$ and $x_{k+1}$ are in $D_{k}$ for $0 \leq k \leq n$. Do a straight-line homotopy between $f$ on $[t_{k}, t_{k+1}]$ and the constant-velocity segment $\overline{x_{k}x_{k+1}}$; this homotopy stays within $D_{k}$, and hence within $\Cpx_{0}^{2}$, because polydisks are convex.

• If the resulting polygonal path, a finite union of segments, avoids the union of the coordinate axes, we're done. Otherwise, use your argument about convex subsets of $\Cpx$ to deform a segment away from the union of the coordinate axes, obtaining a path in $\Cpx_{0}^{2}$. This can be done one coordinate at a time. Alternatively, argue directly that a Cartesian product of path-connected spaces is path-connected, so the product of a disk and a punctured disk, or of two punctured disks, is path-connected. It follows that $D_{k} \cap \Cpx_{0}^{2}$ is path-connected, so we can join $x_{k}$ to $x_{k+1}$ in $D_{k} \cap \Cpx_{0}^{2}$ for each $k$ with $0 \leq k \leq n$. Either way, we get a continuous path between $f(0) = x_{0}$ and $f(1) = x_{n+1}$ in $\Cpx_{0}^{2}$.

Answer 2

Notation: For $(x,y)\in\Bbb C^2$ let $\|(x,y)\|=\sqrt {|x|^2+|y|^2}.$

Let $Y=X\setminus \{(0,0)\}.$ If $(0,0)\not\in X$ then $Y=X$ & the Q is trivial. So assume $(0,0)\in X.$

Take any $b,c\in Y.$ Let $f:[0,1]\to X$ be continuous with $f(0)=b$ and $f(1)=c.$

If $(0,0)\not\in\{f(t): t\in [0,1]\}$ then $f$ is a path in $Y$ from $b$ to $c$ and we are done.

Suppose instead that $0<t_0<1$ and $f(t_0)=(0,0).$ Take $r>0$ such that $2r<\min (\|b\|,\|c\|)$ and such that $\{d\in \Bbb C^2: \|d\|<2r\}\subset X.$ This is possible because $X$ is open and $(0,0)\in X$.

Let $E=\{e\in \Bbb C^2: \|e\|=r\}.$ Then $E\subset Y.$

$(\bullet ). $ Show that $E$ is a path-connected space.

Now let $t_1=\inf \{t\in [0,t_0]:\|f(t)\|\le r \}$ and let $t_2=\sup \{t\in [t_0,1]:\|f(t)\|\le r \}.$ Then $0<t_1\le t_2<1.$ And $f(t_1),f(t_2)\in E.$ And $t\in [0,1]\setminus (t_1,t_2)\implies \|f(t)\|\ge r\implies f(t)\ne (0,0).$

Now $f|_{[0,t_1]}$ is a path in $Y$ from $b$ to $f(t_1)$...

and there is a path in $E\subset Y$ from $f(t_1)$ to $f(t_2)$...

and $f|_{[t_2,1]}$ is a path in $Y$ from $f(t_2)$ to $c.$