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Let be $I, J$ any intervals of the real line and $f: I \longrightarrow J$ a bijection such that $x < y \Longrightarrow f(x) < f(y)$. Prove that $f$ is continuous.

I would like to know if my attempt is correct. Thanks in advance!

My attempt:

We show that $f$ is continuous in $x_0 \in I$. Suppose that $f$ is descontinuous in $x_0 \in I$, this is, $\exists \epsilon > 0$, $\forall \delta > 0;$ $|x - x_0| < \delta$ e $|f(x) - f(x_0)| \geq \epsilon$. Let's look at two cases:

(1) $f(x) \geq f(x_0) + \epsilon$:

As $J$ is an interval, exists $a \in (x_0 - \delta,x_0 + \delta)$ such that $f(a) = f(x_0) + \epsilon$ and exists $c \in J$ tal que $f(x_0) < c < f(a)$, so exists $z \in I$ such that $f(z) = c$ and follows from the fact that $f$ is increasing that $x_0 < z < a$, so $z \in (x_0 - \delta,x_0 + \delta)$, which contradicts the discontinuity of $f$ in $x_0 \in I$.

(2) $f(x) \leq f(x_0) - \epsilon$:

As $J$ is an interval, exists $a \in (x_0 - \delta,x_0 + \delta)$ such that $f(a) = f(x_0) - \epsilon$ and exists $c \in J$ such that $f(a) < c < f(x_0)$, so exists $z \in I$ tal que $f(z) = c$ and follows from the fact that $f$ is increasing that $a < z < x_0$, so $z \in (x_0 - \delta,x_0 + \delta)$, which contradicts the discontinuity of $f$ in $x_0 \in I$.

Therefore $f$ is continuous in $x_0 \in I$.

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  • $\begingroup$ It is correct, but it seems that you have a simple direct argument and transformed it into a contradiction argument. Logically correct, but perhaps a little weird. $\endgroup$ – Darío G Sep 7 '16 at 16:50
  • $\begingroup$ You need only mention surjectivity, as strict monotone increasing generally implies that $F$ is an injection and monotone increasing. I think $\endgroup$ – William Balthes Jun 27 '17 at 7:38
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I would show continuity as follows:

Fix $x_0\in I$ and show that $f$ is continuous at $x_0$.

Let $\epsilon>0$, and we may assume $\epsilon$ small enough so that $[f(x_0)-\epsilon,f(x_0)+\epsilon]\subseteq J$.

Since $f$ is a bijection, there are elements $a,b$ such that $f(a)=f(x_0)-\epsilon$ and $f(b)=f(x_0)+\epsilon$, with $a,b\neq x_0$ (here I am using surjectivity and injectivity of $f$)

Now, since $f$ is increasing and $f(a)=f(x_0)-\epsilon < f(x_0)$, we must have $a<x_0$. Similarly, $x_0<b$.

Let $\delta=\min\{x_0-a,b-x_0\}>0$. Then, for any $x\in (x_0-\delta,x_0+\delta)\subseteq (a,b)$, we have

$$f(a)=f(x_0)-\epsilon <f(x)<f(x_0)+\epsilon=f(b).$$

So, $f$ is continuous at $x_0$.

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  • $\begingroup$ Is this valid? Because the definition say that a function $f$ is continuous in $x_0$ if $\forall \epsilon > 0, \exists \delta > 0; |x - x_0| < \delta \Longrightarrow |f(x) - f(x_0)| < \epsilon$ and you didn't consider any $\epsilon$. $\endgroup$ – George Sep 7 '16 at 22:51
  • $\begingroup$ @George At the beginning of the proof I chose $\epsilon>0$ arbitrary. (that is, $\forall \epsilon>0$.) Then I selected a convenient $\delta>0$ ($\exists \delta>0$). Finally, notice that $x\in (x_0-\delta,x_0+\delta)$ is equivalent to $|x-x_0|<\delta$, and $f(x_0)-\epsilon < f(x) < f(x_0)+\epsilon$ is equivalent to $|f(x)-f(x_0)|<\epsilon$. $\endgroup$ – Darío G Sep 8 '16 at 9:29
  • $\begingroup$ but you said "Let $\epsilon>0$ and we may assume $\epsilon$ small enough so that $[f(x_0)−ϵ,f(x_0)+ϵ] \subset J$.", so your proof isn't is for all $\epsilon$, is it? $\endgroup$ – George Sep 8 '16 at 11:20
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    $\begingroup$ Clearly if $\epsilon$ is bigger than expedted, you can take a smaller $\epsilon'>0$ that works, because $|f(x_0)-f(x)|<\epsilon'$ will imply $|f(x)-f(x_0)|<\epsilon$. Note that in the definition of continuity assuming that $\epsilon$ is "relatively small" is not a problem. The actual problems appear when $\epsilon$ is approaching to zero. $\endgroup$ – Darío G Sep 8 '16 at 12:02
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    $\begingroup$ No. If for example you take $\epsilon=2$ and you can obtain $\delta>0$ such that $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\frac{1}{2}:=\epsilon'$, then in particular you also got $|f(x)-f(x_0)|<2=\epsilon$. $\endgroup$ – Darío G Sep 8 '16 at 12:45

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