# Can we prove $ℝ^n ∖ \{\text{straight line}\}$ and $ℝ^n ∖ \text{\{an axis\}}$ homeomorphic by proving their complements homeomorphic?

I'm trying to prove that $$\mathbb R^n \setminus \{\text{straight line}\}$$ is homeomorphic to $$\mathbb R^n\setminus\text{\{an axis\}}$$.

Their complements, a straight line and an axis, are isomorphic, since the bijective map we find is just a translation from one line to the other. So I thought it could be a good idea to prove that $$G\cong H \Leftrightarrow G^c\cong H^c$$, but I'm not sure that this is generally true.

If it is true, which map could we take to prove it? And if it is false, how should I say that $$\mathbb R^n \setminus \{\text{straight line}\} \cong \mathbb R^n \setminus \{\text{axis}\}?$$ Or at the very least, that $$\mathbb R^n \setminus \{\text{straight line}\} \simeq \mathbb R^n \setminus \{\text{axis}\}$$?.

• That is not true. knots in $\mathbb R^3$ are all homeomorphic, but their complement are not. – Arctic Char Nov 21 '20 at 9:03
• The statement you are trying to prove is false. $\mathbb R^2$ minus a line has two connected components, whereas $\mathbb R^2$ minus an axis has one. – Daniel Mroz Nov 21 '20 at 9:09
• @DanielMroz Then what's an axis? I considered it to be something like the $x$-axis in $\mathbb R^2$, which is a line, namely $\{(x,y):y=0\}$. – Christoph Nov 21 '20 at 17:35
• @Christoph When I wrote my comment I took an axis to refer to a ray: e.g. the (positive) $x$-axis in the plane is $\{(x,0) : x \in \mathbb R ^+ \}$. I don't think this is a terribly irregular usage of the word (we often talk about branch cuts on the positive real axis, for example). But it seems that the intended meaning of an axis in the question is, as you say, the full line. – Daniel Mroz Nov 21 '20 at 17:42

The statement about complements that you claim is not true. First of all, not all topological spaces embed into $$\mathbb R^n$$. Even for those that do, we can find counterexamples. For instance, in $$\mathbb R^2$$ the subsets $$A = \{(x, 0) : x \in (0, 1)\}$$ and $$B = \{(x, 0) : x \in \mathbb R\}$$ are homeomorphic. However, $$\mathbb R^2 - A$$ is connected but $$\mathbb R^2 - B$$ is disconnected. Thus, they are not homeomorphic, so the proof method you had in mind will not work.
Here's a proof for the statement at hand. Let $$L = \{u + tv : t \in \mathbb R\}$$ for some $$u,v \in \mathbb R^n$$ and $$v \neq 0$$ be a line in $$\mathbb R^n$$ and $$A = \{(t, 0, \dots, 0) : t \in \mathbb R\}$$ be an axis. We define a homeomorphism $$f: \mathbb R^n \longrightarrow \mathbb R^n$$ such that $$f[A] = L$$. Indeed, extend $$\{v\}$$ to a basis $$v=v_1, v_2, \dots, v_n$$. We define $$f(x_1, \dots, x_n) = u + \sum x_i v_i$$. This is clearly continuous with a continuous inverse (they are both affine linear isomorphisms). Furthermore, $$f(t, 0,\dots, 0) = u + tv \in L$$ so $$f[A] = L$$ and $$f$$ restricts to a homeomorphism $$f: \mathbb R^n - A \longrightarrow \mathbb R^n - L$$.
In $$\mathbb R$$, $$(0,1)$$ is homeomorphic to $$(1,\infty)$$ (via $$x \to \frac 1 x)$$ but the complement of the first one is not connected whereas the complement of the second interval is connected.