How to prove this subsets of $\mathbb{R}^n$ are homeomorphic to $\mathbb{R}^n$?

Question

Let $f:\mathbb{R}^n\to\mathbb{R}$ be a nonzero linear map. Let $\alpha \in \mathbb{R}$ e define $A=\{x\in \mathbb{R}^n:f(x)<\alpha\}$. By continuity of $f$, $A$ is open in $\mathbb{R}^n$.

How can I prove (rigorously) that $A$ and $\mathbb{R}^n$ are homeomorphic ?

I have some thoughts but I'm not able to make them rigorous. Intuitively I see that $\{x\in \mathbb{R}^n:f(x)=\alpha\}$ is an hyperplane in $\mathbb{R}^n$ which divides $\mathbb{R}^n$ into to "open rectangles" homeomorphics between them, and one of this "rectangles" is $A$.

For example, if $n=2$, I think I can show that $A$ is homeomorphic with $\mathbb{R}\times(-\infty,\alpha)$ maybe using some rotation, and then I easily can show that $\mathbb{R}\times(-\infty,\alpha)$ is homeomrphic to $\mathbb{R}^2$ since I know that $(-\infty,\alpha)$ is homeomrphic to $\mathbb{R}$. For example if $(a,b)\in \mathbb{R}^2$ is the vector such that $f(x,y)=ax+by$, such homeomorphism could be the linear map $\mathbb{R}\times(-\infty,\alpha)\to A$ represented by the matrix \begin{pmatrix} \cos(b/a)& -\sin(b/a) \\\sin(p/a) & \cos(b/a) \end{pmatrix}

What is a rigorous (and possibly elegant) way to prove that $A$ and $\mathbb{R}^n$ are homeomorphic for general $n$?

Suppose also that $\beta\in \mathbb{R}$ and $\beta<\alpha$. Let $B=\{x\in \mathbb{R}^n:f(x)>\beta\}$.

How can I prove that $A \cap B$ is non empty and homoemorphic to $\mathbb{R}^n$ ?

Answer 1

$A$ is a halfspace. Without loss of generality, you can consider $f(x)=x_1$ (with a change of basis, considering a basis completion of $f$, and then looking at the dual basis). Then, it reduces to showing that $(-\infty,a)$ is homeomorphic to $\mathbb{R}$.

Answer 2

Since $f$ is non-zero, $ker(f)$ is an $(n-1)$-dimensional subspace of $\mathbb{R}^n$. Choose any basis $b_1,\dots,b_{n-1}$ of $ker(f)$ and any $b_n$ such that $f(b_n) =1$ (which is possible because $f$ is non-zero linear map). Then $b_1,\dots,b_n$ are linearly independent, thus form a basis of $\mathbb{R}^n$.

The unique linear map $\phi : \mathbb{R}^n \to \mathbb{R}^n$ such that $\phi(e_i) = b_i$, where $e_1,\dots,e_n$ is the standard basis of $\mathbb{R}^n$, is a linear isomorphism and hence a homeomorphism (since linear maps are continuous).

Let $x \in \mathbb{R}$. Write $x = (x',t)$ with $x' \in \mathbb{R}^{n-1}$ and $t \in \mathbb{R}$. Then $\phi(x',t) = \phi(x',0) + \phi(0,t)$, where $\phi(x',0) \in ker(f)$ and $\phi(0,t) = \phi(te_n) = tb_n$. Hence $f(\phi(x',t)) = t$. This shows that for any $J \subset \mathbb{R}$

$$\phi(\mathbb{R}^{n-1} \times J) = \{ x \in \mathbb{R}^n \mid f(x) \in J \} .$$

Therefore $A(\alpha) = \phi(\mathbb{R}^{n-1} \times (\alpha,\infty))$. Similarly, with $A(\alpha,\beta) = \{x \in \mathbb{R}^n \mid \beta < f(x) < \alpha \}$, $A(\alpha,\beta) = \phi(\mathbb{R}^{n-1} \times (\beta,\alpha))$. Since $(\alpha,\infty)$ and $(\beta,\alpha)$ are homeomorphic to $\mathbb{R}$, we are done.

Answer 3

Let $f$ be represented by $a = (a_1, \ldots, a_n) \in \mathbb{R}^n$, $a \ne 0$ via the Riesz representation theorem. WLOG we can assume $a_1 \ne 0$ and by considering $\frac{\alpha}{\|a\|}$, we can assume $\|a\| = 1$.

Let $U : \mathbb{R}^n \to \mathbb{R}^n$ be a unitary map such that $Ua = e_1$. Then $$A = \{x \in \mathbb{R}^n : \langle x,a\rangle < \alpha\} = \{x \in \mathbb{R}^n : \langle Ux,Ua\rangle < \alpha\} = \{x \in \mathbb{R}^n : \langle Ux,e_1\rangle < \alpha\}$$

Consider $\phi : \mathbb{R}^n \to A$ given by $$\phi(x_1, \ldots, x_n) = U^*(\alpha - e^{\langle x, a\rangle}, x_2, \ldots, x_n) \in A$$ Clearly $\phi$ is continuous and $\phi^{-1}$ is given by $$\phi^{-1}(y) = \left(\frac1{a_1}\left(\ln(\alpha - \langle Uy, e_1\rangle) - \sum_{j=2}^n a_j\langle Uy, e_j\rangle\right), \langle Uy, e_2\rangle, \ldots, \langle Uy, e_n\rangle\right)$$ which is also continuous.

If you're unsure about this, explicit calculation gives:

$$\phi^{-1}(\phi(x)) = \phi^{-1}(U^*(\alpha - e^{\langle x, a\rangle}, x_2, \ldots, x_n)) = \left(\frac1{a_1}\left(\langle x, a_1\rangle - \sum_{j=2}^n a_jx_j\right), x_2, \ldots, x_n\right) = x$$ \begin{align} \phi(\phi^{-1}(y)) &= \phi\left(\frac1{a_1}\left(\ln(\alpha - \langle Uy, e_1\rangle) - \sum_{j=2}^n a_j\langle Uy, e_j\rangle\right), \langle Uy, e_2\rangle, \ldots, \langle Uy, e_n\rangle\right) \\ &= U^*\left(\alpha-e^\left(a_1\cdot\frac1{a_1}\left(\ln(\alpha - \langle Uy, e_1\rangle) - \sum_{j=2}^n a_j\langle Uy, e_j\rangle\right) + \sum_{j=2}^na_j\langle Uy, e_n\rangle\right), \langle Uy, e_2\rangle, \ldots, \langle Uy, e_n\rangle\right)\\ &=U^*\left(\alpha - e^{\ln(\alpha- \langle Uy, e_1\rangle)},\langle Uy, e_2\rangle, \ldots, \langle Uy, e_n\rangle\right)\\ &=U^*\left(\langle Uy, e_1\rangle,\langle Uy, e_2\rangle, \ldots, \langle Uy, e_n\rangle\right)\\ &= U^*Uy\\ &= y \end{align}

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In the case with $\beta < \alpha$, we have that $A \cap B = \{x \in \mathbb{R}^n : \langle x, a\rangle \in (\alpha, \beta)\}$. This is nonempty because if $\langle x, a\rangle \ne 0$, then $$\frac{\alpha + \beta}2 \frac{x}{\langle x,a\rangle} \in A \cap B$$ We know that $(\alpha, \beta) \cong \mathbb{R}$. To construct $\phi$ in the preceding proof I used the homeomorphism $x\mapsto \alpha - e^x$ between $\mathbb{R}$ and $(-\infty, \alpha)$.

In this case pick a homeomorphism $g : \mathbb{R} \to (\alpha, \beta)$ and define $\phi : \mathbb{R}^n \to A \cap B$ as $$\phi(x) = U^*(g(\langle x, a\rangle), x_2, \ldots, x_n)$$ and $\phi^{-1}$ similarly.