How do I prove the $f:X\to I$ by $f(x)=\min\{f_i(x_{\beta_i}):i=1,2,...,n\}$ is continuous.

Question

How do I prove that $f:X\to I$ by $f(x)=\min\{f_i(x_{\beta}):i=1,2,...,n\}$ is continuous.

My attempt:- Minimum of two continuous real-valued functions are continuous. $$f_i \circ \pi_{\beta_i}: X\to I$$ be continuous function.So, $f(x)$ can be re-defined as $f(x)=\min\{f_i(x_{\beta_i}):i=1,2,...,n\}=\min\{f_i\circ \pi_{\beta_i}(x):i=1,2,...,n\}$.By Induction, We can induct that minimum of the $n$ continuous real-valued function is continuous.(Where $\pi_{\beta_i}$ is a projection map from $X$ to $X_{\beta_i}$.) Am I correct?

Answer 1

This is again correct. In your first post you proved that the minimum of two continuous real-valued functions is continuous. Induction starts with $n = 1$ and we know that $\min(f_1,\dots,f_{n+1}) = \min(g,f_{n+1})$ where $g = \min(f_1,\dots,f_n)$.