Let be strictly increasing on a subset of . If is open. Prove that is continuous on .

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- Aug 10th 2009, 07:27 PMKat-Mincreasing function
Let be strictly increasing on a subset of . If is open. Prove that is continuous on .

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

Let and be given. Then is in which is open and so there exists an such that .

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 whenever . We already found an epsilon neighborhood of , so we want to construct some points in 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 , be a (weakly) increasing sequence with , and be a (weakly) decreasing sequence with .

The images and of these two sequences are (weakly) increasing and decreasing in , respectively; and both are bounded by , therefore the limits and both exist and we have .

But the assumption that or contradicts openness of combined with monotonicity of f. (Why?) Therefore , which means that f is continuous at x.

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

But the assumption that or contradicts openness of combined with monotonicity of f. (Why?) Therefore , which means that f is continuous at x.

Since was arbitrary, continuity of f on the whole of S follows.

- Aug 11th 2009, 07:45 AMFailure
Well, suppose for example that . If this were the case, then because of and the openness of there would exist an with . By monotonicity of f it follows that for all n, and therefore we get the contradiction (which is to say ).

So,*cannot*be true, and since we know that , the only remaining possiblity is that .

An almost identical argument shows that . - Aug 11th 2009, 07:46 AMKat-Mopen setQuote:

Let and be given. Then is in which is open and so there exists an such that

- Aug 11th 2009, 10:18 AMFailure
Suppose that is given. You seem to agree that because and is open, that there exists at least one such that .

Your only worry seems to be that might not be true. Indeed, that might be a problem. But any that satisfies also has the property that , simply because .

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

- Aug 11th 2009, 08:04 PMFailure
In there is always an infinite number of open intervalls around any given number. For example, all the intervalls are open and centered around the same number . 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 . But once one such open intervall, centered around a given number, is completely contained in that given open set , say, then an infinite number of smaller open intervalls, all centered around the given number, are also contained in that given set, .