Problem 2.19:

If $f$ is a continuous function whose domain includes a closed interval $[a,b]$ an $p \in [a,b]$, then the set of all numbers $x$ \in [a,b]$ such that $f(x) = f(p)$ is a closed point set.

**Proof:**

Let $f$ be a function with domain $D$.

Let $S = [a_{S},b_{S}] \in D$.

Let there be a point $p \in S$.

Let there be a point set $Q = \{m|m = (x,f(x)) = (x,f(p))\}$ such that $Q \in S$.

The set $Q$ is bounded by $S$.

If $Q$ is a closed point set then $Q$ has a left most point and a right most point.

Since $Q$ is bounded by $S$, then the left most point of $Q$ can not be less then the left most point of $S$.

Thus the left most point of $Q$, $a_{Q} \geq a_{S}$.

By symetrical arguement the same can be said for the right most point of $Q$.

This shows that if $a_{Q} = a_{S}$ and $b_{Q} = b_{S}$ then $Q$ is closed.

For the case of $a_{Q} > a_{S}$ or $b_{Q} < b_{S}$, then since $f$ is continuous there are no

points $f(x) < a_{Q}$ where $f(x) \in Q$ thus $a_{Q}$ is the left most point of $Q$.

By symetrical arguement, $b_{Q}$ has no points greater within $Q$.

$\therefore Q$ is a closed point set.

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