Let be strictly increasing on a subset of . If is open. Prove that is continuous on .
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:
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.
why assuming or contradicts the openness and monotonisity?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.
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 .
is open so i understand that for every point of , there is an open set that lies in . but how do we know that the radius is less than the given ? i thought what we know is there is at least one open set centered at with some radius that lies completely in . i didnt know we could pick any radius we wantLet and be given. Then is in which is open and so there exists an such that
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).
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, .