# Thread: topology, metric spaces, proving continuity

1. ## topology, metric spaces, proving continuity

After seeing all the other topology problems be left out to dry, I'm not exactly confident Ill get help, but ill try anyway.

So heres a photo of my problem, since im not the best at latex

ok i get that we have to start with let $\epsilon$ > 0 then we should have $\delta$ > 0 as well right? well after that im seriously sunk, cause I dont know what we would use as as I: (X, d*). I assume whatever we use as that is the one we set < $\delta$ then since its a function to (R,d) it must have |t-a|< $\epsilon$ but even if all that is right, i wouldnt know what to do there. Any help would be amazing

2. ## Re: topology, metric spaces, proving continuity

I think it will really help you to write out what it means for $I$ to be continuous.

To write math you have to use the "tex" tags. Or if you use the advanced posting (there's a button), just click on the [\tex]\sigma[tex] button to add the tags. Here is how some math might look like

Use

d^*(f,g) = \int_a^b |f(t)-g(t)|\; dt for $d^*(f,g) = \int_a^b |f(t)-g(t)|\; dt$
I: (X,d^*) \to (R,d) for $I: (X,d^*) \to (R,d)$.

Just to speed things up somewhat, I've written what it means for $I: (X,d^*) \to (R,d)$ to be continuous. Only look at the spoiler if you are stuck at this step.

Spoiler:
For every $\epsilon >0$, there exists a $\delta > 0$ such that for $f,g\in X$

$d^*(f,g)<\delta$ implies that $d(I(f),I(g)) < \epsilon$.

Now, you want to unravel what $d$ and $d^*$ mean and maybe you'll then see what to do.

3. ## Re: topology, metric spaces, proving continuity

In general, functions are continuous if the inverse image of any open set is open. In metric spaces specifically, the open sets are defined as arbitrary unions of open balls. So since what you're being asked to show is that I is continuous, what you need to show is that I-inverse of any open interval (ball) in R is open in (X, d*). How do you do that?

The standard way is this: You take any element in the inverse image of that open interval, and show that it's contained in an open ball in (X, d*) that is itself contained inside that inverse image. That does it. (Why does that work? See * at bottom)

So... pick any open interval U in R, pick an element f in I-inverse of that U, *find* an open ball centered at f - meaning find an r>0 - so that B(f, r) is a subset of I-inverse of that U, and then you're done. (Why are you done? See ** at bottom)

Blah blah blah abstract. When things get confusing, try to take a step bacwards on the concrete-abstract scale to hope to make them clearer. So I'll work out a more concrete example. To prove that I is continuous, you need to prove that the inverse image of *any* open interval is open. I won't do that. I'll just do it for ONE open interval, to hopefully make clear to you how you might proceed to do it for any open interval.

Let U = (3, 5), a subset of R. Let V = I-inverse of U, a subset of X. I want to show that V is open in (X, d*). Let f in V. I want to find an r>0 such that B(f, r) is a subset of V. But what is V? V is I-inverse of U = {g in X | 3 < I(g) < 5}. Since f in I-inverse of U, we know that 3 < I(f) < 5. Suppose we had found our r. What would that mean? If would mean that B(f, r) a subset of V. And that would mean that if g in B(f, r), then g would also be in V, and so I(g) would be between 3 and 5. In brief, it means that: if d*(f, g) < r, then 3 < I(g) < 5.

So there it is. Given any f such that 3 < I(f) < 5, I need to find r>0 such that: if d*(f, g) < r, then 3 < I(g) < 5.

Now, look at the details of how d* and I are defined. I is linear ( I(cf+g) = cI(f)+I(g) ), and d*(f, g) = I(|f - g|) (Note that if f, g in X, then so are (f-g) and |f-g|). Also, using the properties of integrals, |I(h)| <= I(|h|). Thus |I(f-g)| <= I(|f-g|) = d*(f, g).

So I want r>0 such that I(|f - g|) < r implies 3< I(g) < 5.
But I(|f - g|) < r implies |I(f-g)| < r, so |I(f)-I(g)| < r, so I(f) -r < I(g) < I(f) +r.
Thus if I choose r such that 3 < I(f) -r, and such that I(f) +r < 5, then I can prove that I(|f - g|) < r implies 3< I(f) -r < I(g) < I(f) +r < 5, which is what I need to prove I-inverse of (3,5) is open. But I can choose such an r, since 3 < I(f) < 5. I'll choose r = MIN{ I(f)-3, 5-I(f) }. Note that r>0.

Here it is, put together:

$\text{ Claim: Let } V = I^{-1}( \ (3, 5) \ ). \text{ Then } V \text{ is open in } (X, d^*).$
Proof:

$\text{Choose any } f \in V.$

$\text{ Let } r = min\{ (I(f) - 3), (5 - I(f)) \}. \text{ Note } 3 \le I(f) -r \text{ and } I(f) + r \le 5.$

$\text{Since } f \in V, I(f) \in (3, 5), \text{ and so } 3 < I(f) < 5.$

$\text{ Thus } r = min\{ (I(f) - 3), (5 - I(f)) \} \text{ is positive.}$

$\text{Suppose } g \in B(f, r) \text{ = the open ball of radius } r \text{ centered at } f \text{ in } (X, d^*).$

$\text{ Then } d^*(f, g) < r.$

$\text{Thus } \left\lvert I(f) - I(g) \right\rvert = \left\lvert \int_a^b f - \int_a^b g \right\rvert = \left| \int_a^b (f - g) \right|$

$\le \int_a^b |f - g| = d^*(f, g) < r.$

$\text{So from } \left\lvert I(f) - I(g) \right\rvert < r \text{ have that } -r < I(g) - I(f) < r,$

$\text{ so } I(f) -r < I(g) < I(f) + r,\text{ and so }3 \le I(f) - r < I(g) < I(f) + r \le 5.$

$\text{ Thus } I(g) \in (3, 5), \text{ so } g \in V = I^{-1}((3, 5)).$

$\text{Have shown that if } g \in B(f, r), \text{ then } g \in V. \text{ Therefore } B(f, r) \subset V.$

$\text{So have shown that for any } f \in V, \ \exists \ r>0 \text{ such that } f \in B(f, r) \subset V.$

$\text{Therefore } V = I^{-1}( \ (3, 5) \ ) \text{ is open in } (X, d^*).$

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* Why does that work? Because for each element in the inverse image, you have that it's inside an open ball that's a subset of the inverse image, so that the union of all those balls, for each element in inverse image, is an open set that both contains the inverse image (since every element in the inverse image is in one of those balls), and is also contained in the inverse image (each of those open balls is a subset of the inverse image, so their union is). Thus the union of all those open balls IS the inverse image, and is also open, and hence the inverse image is open.

** Why are you done? Because you've proven that I is continous - because you've proven that I-inverse of that U is open in (X, d*), and since U was arbitrary, every inverse image of an open interval is open in (X, d*), and thus that every that every inverse image of an open set (which is a union of open intervals) is open in (X, d*), and thus that I is continuous.

4. ## Re: topology, metric spaces, proving continuity

We didnt start the ball stuff untill class on the day this stuff was "due" (not really due we just discuss it). But thanks guys you definitely helped me out