Question 6.T.16: If D ⊆ R is compact and f : D → R is continuous and injectiv...
If D ⊆ \mathbb{R} is compact and f : D → \mathbb{R} is continuous and injective, then f^{−1} : f(D) → D is also continuous.
Suppose (y_{n}) is a sequence in f(D) which converges to some y ∈ f(D). It suffices to show that f^{−1}(y_{n}) → f^{−1}(y).
By definition of f(D), there are points x_{n}, x ∈ D such that y_{n} = f(x_{n}) and y = f(x), and we have to show that x_{n} → x. Since D is bounded, the sequence (x_{n}) is bounded. By Theorem 3.15, it suffices to prove that all its convergent subsequences have the same limit x.
Let (x_{n_{k}}) be a convergent subsequence of (x_{n}) with limit x^{\prime} . Since D is closed, x^{\prime} ∈ D, hence
f(x_{n_{k}} ) → f(x^{\prime})
by the continuity of f . But we already have
f(x_{n_{k}}) = y_{n_{k}} → y = f(x).
Therefore f(x^{\prime}) = f(x) by the uniqueness of the limit, and x^{\prime} = x by the injective property of f.