Show that quadrants are homeomorphic Let $V$ be a normed space over $\mathbb{R}$ and $\lambda_1,...,\lambda_n$ be linearly independent continuous linear functionals on $V$.
Define $E^0:=\{x\in V: \forall 1\leq i\leq n, \lambda_i(x)\geq 0\}$ and $E_i:=\{x\in V: \lambda_i(x)\geq 0\}$.
How do I prove that $E^0\cong E_i$ and $E_i\cong E_j$ ? ($\cong$ is a homeomorphic relation)
 A: The linear independence of the $\lambda_i$ implies that the map $T : V \to \mathbb{R}^n$ defined by $T(x) = (\lambda_1(x),\dots,\lambda_n(x))$ is surjective and induces an isomorphism $V/\ker(T) \cong \mathbb{R}^n$. Thus there exists vectors $x_1,\dots,x_n \in V$ such that $\lambda_i(x_j) = \delta_{ij}$, and we can define an isomorphism $\phi : \mathbb{R}^n \times \ker(T) \to V$ by setting $$\phi(a_1,\dots,a_n,x) = a_1 x_1 + \cdots a_n x_n + x.$$ Note that $\phi$ is continuous and that its inverse $\phi^{-1}$ is also continuous since it is given by $$\phi^{-1}(x) = (\lambda_1(x),\dots,\lambda_n(x), x - \sum_{i=1}^n \lambda_i(x) x_i)$$ and the $\lambda_i$ are continuous. This shows that $V$ is homeomorphic to $\mathbb{R}^n \times \ker(T)$.
By restricting $\phi^{-1}$ to $E^0$ and $E_i$, we obtain homeomorphisms $E^0 \cong [0,\infty)^n \times \ker(T)$ and $E_i \cong \mathbb{R}^{n-1} \times [0,\infty) \times \ker(T)$. Hence we only need to show that $[0,\infty)^n$ is homeomorphic to $\mathbb{R}^{n-1} \times [0,\infty)$. This can be done by finding an explicit homeomorphism $[0,\infty)^2 \cong \mathbb{R} \times [0,\infty)$ and then completing the proof by induction.
