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Hi everyone: How would you prove this classic result: if $ n\geq2 ,$ then $\mathbb{C}^{n}\setminus \mathbb{R}^{n}$ is connected? Any reference? Thanks.

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    Hey, apparently [I'm a classic](https://math.stackexchange.com/questions/2091048/can-this-real-set-disconnect-complex-n-space#comment4297918_2091048)! – 2017-01-10
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    @HenningMakholm: Well, of course :) – 2017-01-10
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    Generally $\mathbb R^k\setminus V$ is easily seen to be path-connected if $V$ is a linear subspace of dimension at most $k-2$. Now set $k=2n$. – 2017-01-10
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    @M.Rahmat, please don't ask the same question twice, it's not well received here. – 2017-01-10
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    Hint: Prove first for $m>k$ that a vector space of dimension $m$ over $\mathbb{R}$ minus a subspace of dimension $k$ over $\mathbb{R}$ is homotopy equivalent to $\mathbb{R}^{m-k} - 0$. – 2017-01-10
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    You can prove that it's path connected. – 2017-01-10
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    @AlexM. I think it's safe to say that this is a more specific question; a logical followup to Henning's comment – 2017-01-10

2 Answers 2

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Proof sketch:

  • Prove there is a path between any two points with all nonreal coordinates
  • Prove there is a path from any point to a point with all nonreal coordinates
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    Counterexample: Let $$\omega^{+}_{j}=\lbrace z=x+iy\in\mathbb{C}^{n}:z x,y\in\mathbb{R}^{n},y_{j}>0\rbrace$$ and $$\omega^{-}_{j}=\lbrace z=x+iy\in\mathbb{C}^{n}:z=x+iy, x,y\in\mathbb{R}^{n},y_{j}<0\rbrace$$ for $j=1,2,...,n$. Then set $\Omega_{1}:=\bigcup_{j=1,2,...,n}\omega^{+}_{j}$ and $\Omega_{2}:=\bigcup_{j=1,2,...,n}\omega^{-}_{j}.$ The disjoint open sets $\Omega_{1}$ and $\Omega_{2}$ form a partition of $ \mathbb{C}^{n}\setminus\mathbb{R}^{n} $ and so it is not connected. – 2017-01-13
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    @M.Rahmat $\Omega_1 \cap \Omega_2 \neq \varnothing$. If $j\neq k$, then $\omega_j^+ \cap \omega_k^- \neq\varnothing$. – 2017-01-13
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    @M.Rahmat: Also, in both bullet points, I only put constraints on the endpoints, not on the path itself. The reason for the splitting is to allow one of the, IMO, obvious ways to construct paths: work one coordinate at a time. In the case of the first bullet point, this construction automatically avoids points with all real coordinates. – 2017-01-14
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The set that you are interested in is path connected. Consider $(z_1,\cdots,z_n),(w_1,\cdots,w_n)\in\mathbb{C}^n\setminus\mathbb{R}^n$. Since $(z_1,\cdots,z_n)\in\mathbb{C}^n\setminus\mathbb{R}^n$ there is at least one index $i$ where $z_i$ is not real (some of the other indices can be real, points in $\mathbb{R}^n$ have all coordinates real). Similarly, there is some index $j$ where $w_j$ is not real. We construct two paths. There are two cases:

Case 1: $i\not=j$.

Step 1: While keeping $z_i$ fixed, transform all $z_1,\cdots,z_{i-1},z_{i+1},\cdots,z_n$ into $w_1,\cdots,w_{i-1},w_{i+1},\cdots,w_n$. This can be done with a linear homotopy $(1-t)z_k+tw_k$ changes $z_k$ into $w_k$ as $t$ varies between $0$ and $1$. This never intersects $\mathbb{R}^n$ since the $i$-th coordinate is never real.

Step 2: While keeping all other coordinates fixed, transform $z_i$ into $w_i$. Since the $j$-th coordinate is not real, this path never intersects $\mathbb{R}^n$.

Case 2: $i=j$. In this case, let $k$ be an index other than $i$. Consider the path that keeps all other coordinates fixed and transforms $z_k$ into the complex number $0+1i$ ($i$ is not an index here). Then use Case 1 on $k$ and $j$.

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    Can some explain why the question I asked is "off topic" please? – 2017-01-16
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    Can someone explain why this question "closed as off topic" please? – 2017-01-16