If I remove from $\mathbb{C}^3$ the plane generated by (1,1,0) and (0,0,1) it remains connected (simply connected)? I seen that for $\mathbb{R}^3$ it is not true.
Connected space or not
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general-topology
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0Since $\mathbb C^3$ is homeomorphic to $\mathbb R^6$, while the plane is $4$-dimensional, the answer is yes, because you have plenty of space to move around in – 2017-02-10
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0@5xum So if I remove and the planes generated by {(1,0,1), (0,1,0)} and {(0,1,1), (1,0,0)} it remains simply connected? – 2017-02-10