Show that the subspace $Y = \{ x \in \mathcal{C}[a,b] \mid x(a) = x(b) \} \subset \mathcal{C}[a,b]$ is complete

Question

Now I understand that the function space $\mathcal{C}[a,b]$ of continuous functions from the closed interval $[a,b]$ to $\mathbb{R}$ is complete, with the metric:

$ d(x, y) = \underset{t \ \in \ [a,b]}{\max} \vert x(t) - y(t) \vert $

I also understand the theorem which states that a subspace $M$ of a complete metric space $X$ is itself complete if and only if the set $M$ is closed in $X$.

Finally I understand that $M$ is closed if and only if the situation $x_n \in M, x_n \longrightarrow x$ implies that $x \in M$.

What is don't know what to do is how to put this all together to show that the given subspace $Y$ is closed in $\mathcal{C}[a,b]$.

Answer 1

Let $(x_n)_{n \in \mathbb{N}}$ be a sequence in Y. Suppose that $x_n \rightarrow x.$ We want to prove that $x \in Y$.

For all $\varepsilon>0$, there exists $n_0$ such that $n\geq n_0 \Rightarrow d(x_n,x)< \varepsilon/2$

Since $x_{n_0}(a)=x_{n_0}(b)$, we have: $$|x(a)-x(b)| \leq |x(a)-x_{n_0}(a)|+|x_{n_0}(a)-x_{n_0}(b)|+|x_{n_0}(b)-x(b)| $$ $$=|x(a)-x_{n_0}(a)|+ |x_{n_0}(b)-x(b)| \leq 2d(x_{n_0},x)< \varepsilon $$

Then $|x(a)-x(b)| < \varepsilon$, $\forall \varepsilon>0 \Rightarrow |x(a)-x(b)| = 0 \Rightarrow x(a)=x(b) \Rightarrow x \in Y.$

Thus $Y$ is closed.

Answer 2

Recall that, for $x_n,x \in C[a,b]$, with $d(x_n, x) \to 0$, we have for $t \in [a,b]$: $$ \def\abs#1{\left|#1\right|}\abs{x_n(t) - x(t)} \le d(x_n, x) \to 0 $$ Now suppose $x_n \in Y$, $x_n \to x \in C[a,b]$. Then, by the above $x_n(a) \to x(a)$ and $x_n(b) \to x(b)$. As $x_n(a) = x_n(b)$ for every $n$, we have $$ x(a) = \lim_n x_n(a) = \lim_n x_n(b) = x(b) $$ Hence, $x\in Y$.