increasing function

**Kat-M** · August 10th 2009, 07:27 PM

Let $f$ be strictly increasing on a subset $S$ of $R$ . If $f(S)$ is open. Prove that $f$ is continuous on $S$ .

**siclar** · August 10th 2009, 09:28 PM

Originally Posted by Kat-M

Let $f$ be strictly increasing on a subset $S$ of $R$ . If $f(S)$ is open. Prove that $f$ is continuous on $S$ .

This took a bit of thought, but I figured an epsilon chase works best.

Let $s_0\in S$ and $\epsilon>0$ be given. Then $f(s_0)$ is in $f(S)$ which is open and so there exists an $\epsilon>\gamma>0$ such that $(f(s_0)-\gamma, f(s_0)+\gamma)\subseteq f(S)$ .

Now for the trickier bits, I'll give a few helpful nudges. What we want to show is that there exists a delta such that $|f(s)-f(s_0)|<\epsilon$ whenever $|s-s_0|<\delta$ . We already found an epsilon neighborhood of $f(s_0)$ , so we want to construct some points in $S$ whose images lie in this neighborhood. This is how we want to go about picking our delta.

Big idea 1:

Spoiler:

Big idea 2:

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Big idea 3:

Spoiler:

**Failure** · August 10th 2009, 10:12 PM

Originally Posted by Kat-M

Let $f$ be strictly increasing on a subset $S$ of $R$ . If $f(S)$ is open. Prove that $f$ is continuous on $S$ .

Let $x\in S$ , $\underline{x}_n\in S$ be a (weakly) increasing sequence with $\lim_{n\to \infty}\underline{x}_n=x$ , and $\overline{x}_n\in S$ be a (weakly) decreasing sequence with $\lim_{n\to \infty}\overline{x}_n=x$ .
The images $f(\underline{x}_n)$ and $f(\overline{x}_n)$ of these two sequences are (weakly) increasing and decreasing in $f(S)$ , respectively; and both are bounded by $f(x)$ , therefore the limits $\underline{y}:=\lim_{n\to \infty}f(\underline{x}_n)$ and $\overline{y}:=\lim_{n\to \infty}f(\overline{x}_n)$ both exist and we have $\underline{y}\leq f(x)\leq \overline{y}$ .

But the assumption that $\underline{y}<f(x)$ or $f(x)<\overline{y}$ contradicts openness of $f(S)$ combined with monotonicity of f. (Why?) Therefore $\underline{y}=\lim_{n\to\infty}f(\underline{x_n})= f(x)=\lim_{n\to\infty}f(\overline{x}_n)=\overline{ y}$ , which means that f is continuous at x.

Since $x\in S$ was arbitrary, continuity of f on the whole of S follows.

**Kat-M** · August 11th 2009, 07:30 AM

But the assumption that $\underline{y}<f(x)$ or $f(x)<\overline{y}$ contradicts openness of $f(S)$ combined with monotonicity of f. (Why?) Therefore $\underline{y}=\lim_{n\to\infty}f(\underline{x_n})= f(x)=\lim_{n\to\infty}f(\overline{x}_n)=\overline{ y}$ , which means that f is continuous at x.

Since $x\in S$ was arbitrary, continuity of f on the whole of S follows.

why assuming $\underline{y}<f(x)$ or $f(x)<\overline{y}$ contradicts the openness and monotonisity?

**Failure** · August 11th 2009, 07:45 AM

Originally Posted by Kat-M

why assuming $\underline{y}<f(x)$ or $f(x)<\overline{y}$ contradicts the openness and monotonisity?

Well, suppose for example that $\underline{y}<f(x)$ . If this were the case, then because of $f(x)\in f(S)$ and the openness of $f(S)$ there would exist an $x_0\in S$ with $\underline{y}<f(x_0)<f(x)$ . By monotonicity of f it follows that $\underline{x}_n< x_0 < x$ for all n, and therefore we get the contradiction $x=\lim_{n\to\infty}\underline{x}_n\leq x_0<x$ (which is to say $x<x$ ).

So, $\underline{y}< f(x)$ cannot be true, and since we know that $\underline{y}\leq f(x)$ , the only remaining possiblity is that $\underline{y}=f(x)$ .
An almost identical argument shows that $f(x)=\overline{y}$ .

**Kat-M** · August 11th 2009, 07:46 AM

Let $s_0\in S$ and $\epsilon>0$ be given. Then $f(s_0)$ is in $f(S)$ which is open and so there exists an $\epsilon>\gamma>0$ such that $(f(s_0)-\gamma, f(s_0)+\gamma)\subseteq f(S)$

$f(S)$ is open so i understand that for every point $f(s_0)$ of $f(S)$ , there is an open set that lies in $f(S)$ . but how do we know that the radius $\gamma$ is less than the given $\epsilon$ ? i thought what we know is there is at least one open set centered at $f(s_0)$ with some radius that lies completely in $f(S)$ . i didnt know we could pick any radius we want

**Failure** · August 11th 2009, 10:18 AM

Originally Posted by Kat-M

$f(S)$ is open so i understand that for every point $f(s_0)$ of $f(S)$ , there is an open set that lies in $f(S)$ . but how do we know that the radius $\gamma$ is less than the given $\epsilon$ ? i thought what we know is there is at least one open set centered at $f(s_0)$ with some radius that lies completely in $f(S)$ . i didnt know we could pick any radius we want

Suppose that $\varepsilon >0$ is given. You seem to agree that because $f(s_0)\in f(S)$ and $f(S)$ is open, that there exists at least one $\gamma >0$ such that $(f(s_0)-\gamma,f(s_0)+\gamma)\subseteq f(S)$ .

Your only worry seems to be that $\varepsilon>\gamma>0$ might not be true. Indeed, that might be a problem. But any $\gamma'$ that satisfies $0<\gamma'\leq \gamma$ also has the property that $(f(s_0)-\gamma',f(s_0)+\gamma')\subseteq f(S)$ , simply because $(f(s_0)-\gamma',f(s_0)+\gamma')\subseteq (f(s_0)-\gamma,f(s_0)+\gamma)\subseteq f(S)$ .
Thus, the possibility that our initial value for $\gamma$ might in fact not be smaller than the given $\varepsilon$ is not a particularly big worry, because we can always replace it with $\min(\gamma,\varepsilon/2)$ if need be. Indeed, we would do this in a "preventive" fashion, just to make sure $\varepsilon>\gamma>0$ holds (afterwards).

**Kat-M** · August 11th 2009, 01:26 PM

So if there is an open set centered at $s_0$ with radius $\gamma$ , then any $\gamma ' \leq \gamma$ can be a radius of an open set around $s_0$ ? which means that there are infinitely many open sets with radius less than $\gamma$ .

**Failure** · August 11th 2009, 08:04 PM

Originally Posted by Kat-M

So if there is an open set centered at $s_0$ with radius $\gamma$ , then any $\gamma ' \leq \gamma$ can be a radius of an open set around $s_0$ ? which means that there are infinitely many open sets with radius less than $\gamma$ .

In $\mathbb{R}$ there is always an infinite number of open intervalls around any given number. For example, all the intervalls $(f(s_0)-\tfrac{1}{n},f(s_0)+\tfrac{1}{n})$ are open and centered around the same number $f(s_0)$ . They can get arbitrarily small - and yet they are all open.
The particular problem in your case was that those intervalls had to be contained in a given open set $f(S)$ . But once one such open intervall, centered around a given number, is completely contained in that given open set $f(S)$ , say, then an infinite number of smaller open intervalls, all centered around the given number, are also contained in that given set, $f(S)$ .

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