Let $f: \mathbb{R}^m \rightarrow \mathbb{R}$. Show that $f(x) = x_i$ where $x_i$ is the $i^{nd}$ component of $x$ regarding the standard basis of $\mathbb{R}^m$ is continuous in 0.
The definition says :
Let $A \subset \mathbb{R}^m$, $f: A \rightarrow \mathbb{R}^p$ and $a \in A$. The function is continuous in a when for all $\epsilon \gt 0$, $\exists \delta \gt 0$ such that for all $x \in A$, $||x - a|| \le \delta$ and $|| f(x) - f(a) || \le \epsilon$.
I know $f(x) = (e_i|x)$.
By Cauchy-Schwarz inequality, we have : $|(e_i|x)| \le ||e_i||.||x||$
We have $||x - 0|| = ||x|| \le \delta$.
So $||f(x) - f(0)|| \le ||e_i||.||x|| - 0 = ||e_i||.||x||$
Now, for every $\epsilon \gt 0$ we can take $\delta = \frac{\epsilon}{||e_i||}$
So $||f(x) - f(0)|| \le ||e_i||.||x|| \le ||e_i||\delta = ||e_i||\frac{\epsilon}{||e_i||} = \epsilon$.
I wonder if it's correct ?