This proves the theorem in the case where f is strictly decreasing. The case where f is
increasing is similar. ■

Proof 2: Note that since f is either strictly increasing or strictly decreasing, it maps
an open interval to an open interval. Now let U be an open set. Thus U = ∪_{x∈U}I_{x} where
I_{x} is an open interval. Then

(f −1)−1(U) = (f−1)−1(∪ I) = ∪ (f −1)− 1(I)
( x)∈−U1 x x∈U x
= ∪x∈U f−1 (Ix ∩ (a,b)) = ∪x∈U f (Ix ∩ (a,b))

which is a union of open intervals and is therefore, open. This uses the fact that an
equivalent condition to continuity is that inverse images of open sets are open. This was
not explicity shown above, but you might try to show this. It is close to what was
shown.

As to the last claim, it follows right away from the observation that f

([a,b])

is
compact. Say f

(xn)

→ y. Then since f

([a,b])

is closed and bounded, y ∈ f

([a,b])

. Say
y = f

(x)

. Thus f

(xn)

→ f

(x)

? Does x_{n} = f^{−1}

(f (xn ))

→ f^{−1}

(f (x ))

? If not, there
exists a subsequence, still denoted as

{xn}

and ε > 0 such that

|xn − x|

≥ ε. Now
there is a further subsequence, still denoted as