Nearest point map in finite-dimensional subspace of normed space is continuous I am going through Carothers's A Short Course on Approximation Theory (link), and contemplating the following lemma.

Lemma 1.9. Let $Y$ be a finite-dimensional subspace of a normed linear space $X$, and suppose that each $x \in X$ has a unique nearest point $y_x \in Y$ such that $\displaystyle ||x-y_x|| = \min_{y \in Y} ||x-y||$. Then the nearest point map $P: x \to Y; x \mapsto y_x$ is continuous.


Proof (sketch): Suppose $x_n \to x$ in $X$.

*

*Show $(P(x_n))$ is bounded, and hence has a convergent subsequence (say, to $P_0$).

*By "passing to a subsequence", $P(x_n) \to P_0$.

*Establish the inequality $||x-P_0|| \leq ||x-P(x)||$, and thus $P_0 = P(x)$ by the uniqueness of nearest points.


I have the following questions, corresponding to the steps numbered above:

*

*Is this a straightforward application of Bolzano-Weierstrass?

*Why does the convergence of a subsequence of $(P(x_n))$ imply the convergence of the whole sequence?

 A: *

*First, prove that $(P(x_n))$ is bounded:
$$\begin{align}d(Y,x)&\le\|P(x_n)-x\|\\&\le\|P(x_n)-x_n\|+\|x_n-x\|\\&=d(Y,x_n)+\|x_n-x\|\\&\le d(Y,x)+2\|x_n-x\|.\end{align}$$
Then, "straightforward application of Bolzano-Weierstrass" indeed.

*It is false that "the convergence of a subsequence of $(P(x_n))$ implies the convergence of the whole sequence". Instead, two methods are possible:


*

*the method of Theo's comment: suppose we shall be able to prove (in 3.) that the limit of every convergent subsequence of $(P(x_n))$ is $P(x).$ Using Bolzano-Weierstrass again, this will guarantee that $P(x_n)\to P(x),$ thus ending the proof. So, replacing $(x_n)$ by some subsequence, still denoted by $(x_n)$ from now on, such that $P(x_n)\to P_0$, there remains to prove that $P_0=P(x).$

*the method of Jake Lai's comment, using Carother's exercise 1.10, valid for a map $f:X\to Y$ where $X,Y$ are any metric spaces, and better phrased: If [whenever $x_n\to x$, there exists a subsequence of $(f(x_n))$ converging to $f(x)$], then $f$ is continuous at $x.$ So, there remains to show that $P_0=P(x),$ where $P_0$ is the limit of the converging subsequence of inherited from 1. and still denoted by $(P(x_n))$ from now on.

For the sake of completeness, here is a solution of Carothers's auxiliary exercise 1.10: Assume for contradiction that $f$ is not continuous at $x,$ i.e. there exists a sequence $(x_n)$ converging to $x$ and such that $f(x_n)\not\to f(x).$ Then, for some $\epsilon>0,$ there exists, for all $N\in\Bbb N,$ some $n\ge N$ satisfying $d(f(x_n),f(x))\ge\epsilon.$ This allows to construct a subsequence $(x_{n_k})$ such that $d(f(x_{n_k}),f(x))\ge\epsilon$ for all $k.$ This sequence $(x_{n_k})$ converges to $x$ and there exists no subsequence of $(f(x_{n_k}))$ converging to $f(x),$ thereby ending the proof by contradiction.
