If $f:\mathbb{R}\to\mathbb{R}$ is continuous and bijective. Is $f^{-1}:\mathbb{R}\to\mathbb{R}$ continuous?

Question

I have a function $f:\mathbb{R}\to\mathbb{R}$ that is both continuous and bijective. My instinct tells me that $f^{-1}$ is also continuous.but I'm not able to show it. All I can show is that f is strictly monotone. Can anyone show how to move on from this point onwards to show the continuity of $f^{-1}$. If possible you can use topology for the proof. Thanks in advance

Answer 1

The answer is yes.

Since $f$ is continuous and injective, it is strictly monotone on $\mathbb{R}$. Then $f^{-1}$ is also strictly monotone on $\mathbb{R}$. Continuity of $f^{-1}$ follows from this lemma:

Let $g : \mathbb{R} \to \mathbb{R}$ be a strictly monotone surjection. Then $g$ is continuous on $\mathbb{R}$.

Proof:

WLOG assume that $g$ is strictly increasing. Let $c \in \mathbb{R}$ be arbitrary. We will show that $g$ is continuous at $c$.

Pick an interval $\langle a,b\rangle$ such that $c \in \langle a,b\rangle$. Let $0 < \varepsilon < \min\left\{g\left(\frac{c+b}2\right) - g(c), g(c) - g\left(\frac{a+c}2\right)\right\}$ be arbitrary. We have

$$g(c) < g(c) + \varepsilon < g\left(\frac{c+b}2\right)$$

$g$ is surjective so $\exists \delta' > 0$ such that $g(c + \delta') = g(c) + \varepsilon$.

We have

$$g\left(\frac{a+c}2\right) < g(c) -\varepsilon < g(c)$$

$g$ is surjective so $\exists \delta'' > 0$ such that $g(c - \delta'') = g(c) - \varepsilon$.

Let $\delta = \min\{\delta',\delta''\}$. For $x \in \langle c, c+\delta\rangle$ we have

$$g(c) < g(x) < g(c+\delta) \le g(c+\delta')= g(c) + \varepsilon$$

For $x \in \langle c-\delta, c\rangle$ we have

$$g(c) - \varepsilon < g(c-\delta'') \le g(c- \delta) < g(x) < g(c)$$

Therefore $|x-c| < \delta \implies |g(x) - g(c)| < \varepsilon$ so $g$ is continuous at $c$.

Answer 2

If $f$ is strictly monotone, show it takes intervals to intervals (use the fact it is continuous to show the image is a full interval). Now reason that $f$ is an open map, since every open set is the union of intervals.

Answer 3

Let me show you an approach via the definition of continuity over sequences. For this, first recall:

Definition: Let $f:D\subseteq\mathbb{R}\to\mathbb{R}$ a function. $f$ is continuous in $x_0\in D$ if for every sequence $(x_n)$ in $D$ s.t. $$\lim_{n\to\infty}x_n=x_0$$it holds that $$\lim_{n\to\infty}f(x_n)=f(x_0)$$

Note, that if $f$ is continuous and bijective, then it is strictly monotone(rising or falling).

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You may generalize your result even for functions $f:I\to\mathbb{R}$ for $I\subseteq\mathbb{R}$ an interval. For this, it is useful to recall the following lemma:

Lemma: $I\subseteq\mathbb{R}$ is an interval iff $\forall x,y\in I: x<y$ implies $[x,y]\subseteq I$.

Note also the following:

Lemma: Let $I\subseteq\mathbb{R}$ be an interval and $f:I\to f(I)$ be strictly monotone. Then $f^{-1}$ is strictly monotone as well.

This follows easily, as $f$ is strictly monotone, thus injective and even bijective on $I$ with $f(I)$. Now, we are ready to prove:

Proposition: Let $I\subseteq\mathbb{R}$ be an interval and $f:I\to\mathbb{R}$ continuous on $I$ and strictly monotone. Then $f(I)$ is an interval and $f^{-1}$ is continuous on $f(I)$.

Proof: W.l.o.g. let $f$ be strictly monotone rising(the case for falling is analogous). First, note that $f(I)$ really is an interval. For this, let $x,y\in f(I)$ with $x<y$ and let $z\in [x,y]$. There are $a,b\in I$ s.t. $f(a)=x$ and $f(b)=y$. Thus, we have $$f(a)=x\leq z\leq y=f(b)$$ By the intermediate value theorem, there is a $c\in I$ s.t. $f(c)=z$. Thus, $z=f(c)\in f(I)$ and thus $[x,y]\subseteq f(I)$. By the lemma above, $f(I)$ is an interval.

Now, we show that $f^{-1}$ is continuous on $f(I)$. Let $I$ first be a compact interval, i.e. $I=[x,y]$ for some $x,y\in\mathbb{R}$ with $x<y$. Then, by the first part of the proof, $f(I)=[f(x),f(y)]$ as $f$ is strictly monotone rising.

Now, let $(a_n)$ be a sequence in $f(I)$ s.t. $$\lim_{n\to\infty}a_n=a_0$$ and let $b_n=f^{-1}(a_n)$ f.a. $n\in\mathbb{N}$ and $b_0=f^{-1}(a_0)$. Since $I$ is bounded, by the theorem of Bolzano-Weierstrass, $(b_n)$ has a least one limit point $c$, i.e. there is a subsequence $(b_{n_k})$ converging to $c$.

Since $I$ is compact, we also have $c\in I$. Since $f$ is continuous, we have $$\lim_{k\to\infty}f(b_{n_k})=f(c)$$ But now, $f(b_{n_k})=a_{n_k}$ and thus, as $$\lim_{k\to\infty}a_{n_k}=a_0$$ as well(as the limit is unique), we have $f(c)=a_0=f(b_0)$. Since $f$ is injective, we have $b_0=c$. Thus, $(b_n)$ is convergent with $\lim_{n\to\infty}b_n=b_0$.

We can now generalize this to any interval. Per definition, $f^{-1}$ is continuous on $f(I)$ if and only if $f^{-1}$ is continuous on every point in $f(I)$.

We can thus show that $f^{-1}$ is continuous on every $[x,y]\subseteq f(I)$. As every point can be wrapped in a closed interval, the claim thus follows:

Let $x,y\in f(I)$ with $x<y$. Then again, there are $a,b\in I$ with $f(a)=x$ and $f(b)=y$ where we have $f(a)<f(b)$ as $f$ is strictly monotone rising. Now, $[f(a),f(b)]$ is a compact interval and by the above, we have that $f^{-1}$ is continuous on $[f(a),f(b)]=[x,y]$. $\Box$