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

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- Aug 10th 2009, 07:27 PMKat-Mincreasing function
Let $\displaystyle f$ be strictly increasing on a subset $\displaystyle S$ of $\displaystyle R$. If $\displaystyle f(S)$ is open. Prove that $\displaystyle f$ is continuous on $\displaystyle S$.

- Aug 10th 2009, 09:28 PMsiclar
This took a bit of thought, but I figured an epsilon chase works best.

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

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 $\displaystyle |f(s)-f(s_0)|<\epsilon$ whenever $\displaystyle |s-s_0|<\delta$. We already found an epsilon neighborhood of $\displaystyle f(s_0)$, so we want to construct some points in $\displaystyle S$ whose images lie in this neighborhood. This is how we want to go about picking our delta.

Big idea 1:__Spoiler__:

Big idea 2:__Spoiler__:

Big idea 3:__Spoiler__: - Aug 10th 2009, 10:12 PMFailure
Let $\displaystyle x\in S$, $\displaystyle \underline{x}_n\in S$ be a (weakly) increasing sequence with $\displaystyle \lim_{n\to \infty}\underline{x}_n=x$, and $\displaystyle \overline{x}_n\in S$ be a (weakly) decreasing sequence with $\displaystyle \lim_{n\to \infty}\overline{x}_n=x$.

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

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

Since $\displaystyle x\in S$ was arbitrary, continuity of f on the whole of S follows. - Aug 11th 2009, 07:30 AMKat-MopennessQuote:

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

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

- Aug 11th 2009, 07:45 AMFailure
Well, suppose for example that $\displaystyle \underline{y}<f(x)$. If this were the case, then because of $\displaystyle f(x)\in f(S)$ and the openness of $\displaystyle f(S)$ there would exist an $\displaystyle x_0\in S$ with $\displaystyle \underline{y}<f(x_0)<f(x)$. By monotonicity of f it follows that $\displaystyle \underline{x}_n< x_0 < x$ for all n, and therefore we get the contradiction $\displaystyle x=\lim_{n\to\infty}\underline{x}_n\leq x_0<x$ (which is to say $\displaystyle x<x$).

So, $\displaystyle \underline{y}< f(x)$*cannot*be true, and since we know that $\displaystyle \underline{y}\leq f(x)$, the only remaining possiblity is that $\displaystyle \underline{y}=f(x)$.

An almost identical argument shows that $\displaystyle f(x)=\overline{y}$. - Aug 11th 2009, 07:46 AMKat-Mopen setQuote:

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

- Aug 11th 2009, 10:18 AMFailure
Suppose that $\displaystyle \varepsilon >0$ is given. You seem to agree that because $\displaystyle f(s_0)\in f(S)$ and $\displaystyle f(S)$ is open, that there exists at least one $\displaystyle \gamma >0$ such that $\displaystyle (f(s_0)-\gamma,f(s_0)+\gamma)\subseteq f(S)$.

Your only worry seems to be that $\displaystyle \varepsilon>\gamma>0$ might not be true. Indeed, that might be a problem. But any $\displaystyle \gamma'$ that satisfies $\displaystyle 0<\gamma'\leq \gamma$ also has the property that $\displaystyle (f(s_0)-\gamma',f(s_0)+\gamma')\subseteq f(S)$, simply because $\displaystyle (f(s_0)-\gamma',f(s_0)+\gamma')\subseteq (f(s_0)-\gamma,f(s_0)+\gamma)\subseteq f(S)$.

Thus, the possibility that our initial value for $\displaystyle \gamma$ might in fact not be smaller than the given $\displaystyle \varepsilon$ is not a particularly big worry, because we can always replace it with $\displaystyle \min(\gamma,\varepsilon/2)$ if need be. Indeed, we would do this in a "preventive" fashion, just to make sure $\displaystyle \varepsilon>\gamma>0$ holds (afterwards). - Aug 11th 2009, 01:26 PMKat-Mopen set
So if there is an open set centered at $\displaystyle s_0$ with radius $\displaystyle \gamma$, then any $\displaystyle \gamma ' \leq \gamma$ can be a radius of an open set around $\displaystyle s_0$? which means that there are infinitely many open sets with radius less than $\displaystyle \gamma$.

- Aug 11th 2009, 08:04 PMFailure
In $\displaystyle \mathbb{R}$ there is always an infinite number of open intervalls around any given number. For example, all the intervalls $\displaystyle (f(s_0)-\tfrac{1}{n},f(s_0)+\tfrac{1}{n})$ are open and centered around the same number $\displaystyle f(s_0)$. 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 $\displaystyle f(S)$. But once one such open intervall, centered around a given number, is completely contained in that given open set $\displaystyle f(S)$, say, then an infinite number of smaller open intervalls, all centered around the given number, are also contained in that given set, $\displaystyle f(S)$.