Let $V$ be the span of $(1,1,1)$ and $(0,1,1) \in \Bbb R^3$. Let $u_1=(0,0,1),u_2=(1,1,0)$ and $u_3=(1,0,1)$. Which of the following is/are correct?

$1. \ $ $(\Bbb R^3 \setminus V) \cup \{(0,0,0)\}$ is not connected.

$2. \ $ $(\Bbb R^3 \setminus V) \cup \{tu_1+(1-t)u_3 : 0 \le t \le 1 \}$ is connected.

$3. \ $ $(\Bbb R^3 \setminus V) \cup \{tu_1+(1-t)u_2 : 0 \le t \le 1 \}$ is connected.

$4. \ $ $(\Bbb R^3 \setminus V) \cup \{(t,2t,2t):t \in \Bbb R \}$ is connected.

$(1),(3)$ and $(4)$ are path-connected spaces so they are connected. Hence clearly $(1)$ is false and $(3)$ and $(4)$ are correct options. I think $(2)$ is also path-connected as $(0,0,1) \in V$. So the space is such that it is the three dimensional Euclidean space separated by the plane $V$ which is $x=y$ in such a way that the plane has a fracture along the line segment joining $(1,0,1)$ and $(0,0,1)$ . Now since $(0,0,1) \in V$ we can join any two points lying on two sides by joining them by a curve passing through $(0,0,1)$ not cutting any other point of the plane $V$. Is it not possible?

Please help me in this regard. I am quite confused at this stage about whether my logic is valid or not.

Thank you very much.


1 Answer 1


Your idea and logic are correct: If $V$ is a proper subspace and $S$ a subset of $\Bbb R^n$, then $(\Bbb R^n\setminus V)\cup S$ is connected and path-connected if and only if $S\cap V\ne \emptyset$. However, your result for 2. is false because the starting point is false: $(0,0,1)\notin V$. Note that $y=z$ for all points $(x,y,z)$ in $V$ and $y=0\ne 1=z$ for all points in the line segment.

  • $\begingroup$ Oh! I am too silly. Thanks @Hagen von Eitzen for your help. $\endgroup$
    – D_C
    Commented Jun 8, 2018 at 5:19

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