Showing that a set $D$ is closed and open

Question

I have this theorem in my book:

Suppose that $f$ is continuous in the rectangle $$R = \{(t,u) \mid |t-t_0|\leq T, |u-u_0|\leq L\} $$ and that $$|f(t,u)|\leq M \text{ if } (t,u)\in R$$ Let $\delta = \min(T,L/M).$ If $u(t)$ is any solutions of $\dot{u} = f(t,u)$, $u(t_0) = u_0$ then $$|u(t)-u_0|\leq L \text { when } |t-t_0|\leq \delta.$$

The proof is stated as:

If a solution $u(t)$ stays inside the interval $|u(t)-u_0|\leq L$, then its derivative is bounded by $M$, so the solution cannot escape the interval in less time than $L/M$. Consider $$D = \{0\leq\eta\leq\delta \mid |u(t)-u_0|\leq L \text{ for all } |t-t_0|\leq\eta\}. $$ Then $0\in D$, and if $\eta\in D$, then $\eta ' \in D$ for all $0\leq \eta '\leq \eta$. Thus $D$ is a nonempty interval. Moreover, $D$ is closed in $[0,\delta]$ because $u(t)$ is a continuous function of $t$. If $\eta\in D$ and $\eta<\delta$, then $f(t,u(t))\leq M$ for $|t-t_0|\leq \eta$, so $$|u(t)-u_0|\leq \bigg |\int_{t_0}^t f(s,u(s))ds\bigg |\leq M\eta < M\delta = L.$$ Since we have a strict inequality, and $u$ is continuous, it follows that there is an $\epsilon>0$ such that $|u(t)-u_0|\leq L$ when $|t-t_0|\leq \eta + \epsilon$. Thus, $D$ is open in $[0,\delta]$, from which we conclude that $D = [0,\delta]$.

My questions are:

How does $u(t)$ being a continuous function of $t$ imply $D$ is closed?

Why does the second argument show it is open?

Also if a non-empty set $A$ is open and closed on some interval $[a,b]$ then this implies that $A = [a,b]$? In general?

Any help and comments would be appreciated. Thank you in advance.

Answer 1

How does u(t) being a continuous function of t imply D is closed?

If $\eta_k \in D$ converges to $\eta$ and $|t-t_0| \le \eta$, then we can choose a sequence $t_k \to t$ such that $|t_k - t_0| \le \eta_k$. We then have $$|u(t) - u(t_0)| \le |u(t) - u(t_k)| + |u(t_k) - u(t_0)|.$$ As $k \to \infty$, the first term converges to zero by the continuity of $u$ and the second term is not greater than $L$, so $|u(t) - u(t_0)| \le L$ and thus $\eta \in D$.

Why does the second argument show it is open?

For every $\eta \in D, \eta<\delta$ there is an $\epsilon > 0$ such that $|u(t) - u_0| \le L$ when $|t - t_0| \le \eta + \epsilon$, which means exactly that $\eta + \epsilon \in D$. Since we have already established that $D$ is an interval containing $[0,\eta]$, this means the full neighbourhood $[0,\eta+\epsilon]$ of $\eta$ is also in $D$; i.e. $D$ is open in $[0,\delta]$.

Also if a non-empty set $A$ is open and closed on some interval $[a,b]$ then this implies that $A=[a,b]$? In general?

Yes - in more generality, whenever a set $A \subset X$ is open and closed in a connected space $X$, it is either empty or all of $X$. This is pretty much the definition of connectedness in general topology.