# Is this a typo? $(t_0,x_0)\in \mathbb R^{n+1}$ or $(x_0,t_0)\in \mathbb R^{n+1}$?

Is this a typo?

From Ordinary Differential Equations and Dynamical Systems by Gerald Teschl (free copy), page 36, equation $$2.10$$ says:

Initial value problem \begin{align} \dot x &= f(t,x) \tag 1\\ x(t_0) &= x_0 \tag 2 \end{align} We suppose $$f\in C(U,\mathbb R^n)$$, where $$U$$ is an open subset of $$\mathbb R^{n+1}$$ and $$(t_0,x_0)\in U$$.

Question:

Isn't the order wrong in $$(t_0,x_0)$$ and $$f(t,x)$$ due to $$\mathbb R^{n+1}$$?

For simplicity I here use $$\mathbb R^{n+1}$$ instead of the subset $$U$$. So, from $$(t_0,x_0)\in \mathbb R^{n+1}$$ we have that $$t_0 \in \mathbb R^n$$ and $$x_0 \in \mathbb R$$, but in $$(1)$$ the variable is $$t$$? If $$t\in \mathbb R$$, isn't also $$t_0 \in \mathbb R$$?

I.e. shouldn't it be \begin{align} \dot x &= f(x,t) \tag 3\\ (x_0,t_0) &\in U \tag 4\\ U &\subset \mathbb R^{n+1} \tag 5 \end{align}

Alternatively $$\mathbb R \times \mathbb R^n =\mathbb R^{1+n}$$ for $$(1)-(2)$$.

• But $\mathbb{R}^{1 + n}$ is the same as $\mathbb{R}^{n + 1}$. I think it is clear from the context that $x\in\mathbb{R}^n$ since he said $f\in C(U, \mathbb{R}^n)$. Commented Dec 10, 2020 at 23:58

It sounds like you are viewing $$\mathbb{R}^{n+1}$$ as $$\mathbb{R}^{n}\times\mathbb{R}$$.
But $$\mathbb{R}^{n+1}$$ is also $$\mathbb{R}\times \mathbb{R}^{n}$$.
The integer $$n+1$$ is the same as the integer $$1+n$$.