# Prove A~B=>f(A)~f(B) for a continuous f:X->Y

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1. Nov 17, 2015

### BiGyElLoWhAt

So proofs are a weak point of mine.
The hint is that a composite of a continuous function is continuous. I'm not really sure how to use that. What I was thinking was something to the effect of an epsilon delta proof, is that applicable?

Something to the effect of:
$A \sim B\text{ and let } f \text{ be a continuous function X} \to \text{Y:}$
$\text{by definition, if } f \text{ is continuous, then there exists an } \epsilon \text{ for every } \delta \text{ such that ... blah...so }$
$f(A) \sim f(A+\epsilon) \sim f(A+n\epsilon) \text{ and by induction } f(A) \sim f(B) \text{ for large enough n}$
is that strong enough? Since A+n*epsilon = B it goes to f(B). Would that be necessary in the proof?
How can I do a proof with composite continuity?

2. Nov 17, 2015

### micromass

You will have to say what $X$ and $Y$ are, and what $\sim$ is.

3. Nov 17, 2015

### Krylov

At first I thought I was going to make an effort to understand the OP, but I just gave up.

4. Nov 17, 2015

### micromass

My bet is that the OP means that if two numbers $A$ and $B$ are infinitesimally close, then $f(A)$ and $f(B)$ are also infinitesimally close. This is indeed a characterization of continuity in the hyperreal number system.

5. Nov 17, 2015

### Krylov

Both your bet and such characterisation sound plausible. I just experienced a parsing error when I encountered the "... blah...so".

6. Nov 17, 2015

### micromass

My parsing error came a bit earlier with "there is an epsilon for each delta"

7. Nov 17, 2015

### Krylov

I remember a mischievous "true/false" question on an introductory analysis exam, where the definition of continuity was stated as usual, but with the roles of the symbols $\varepsilon$ and $\delta$ reversed, to the dismay of the audience.

8. Nov 17, 2015

### micromass

You mean the following?
$$\forall \delta >0: \exists \varepsilon>0: \forall x: |x-a|<\varepsilon~\Rightarrow~|f(x) - f(a)| < \delta$$
That's a really good question :D

9. Nov 17, 2015

### Krylov

Yes, precisely.

10. Nov 17, 2015

### micromass

Could be worse:
A function $x:\mathbb{R}\rightarrow \mathbb{R}$ is continuous at $\delta$ if and only if
$$\forall a>0:~\exists \varepsilon>0: \forall f\in \mathbb{R}:~|f-\delta|<\varepsilon~\Rightarrow~|x(f)-x(\delta)|<a$$

11. Nov 17, 2015

### Staff: Mentor

12. Nov 17, 2015

### BiGyElLoWhAt

Terribly sorry, all.
http://inperc.com/wiki/index.php?title=Homology_classes
There's where it comes from.
It isn't defined anywhere, but I'm assuming ~ means connected, no cuts/holes/ etc.
X is a subspace of R^n, and I'm assuming Y is as well, X is the only one that has actually been defined, but thus far, as has been given, we're only working in subspaces of R^n.

Does that clear things up? The Blah... is the usual definition for continuity in calculus, I was just being lazy.

*The exercise is about 1/3 of the way down, just before section 3 about counting features.

13. Nov 17, 2015

### micromass

So $A\sim B$ means that there is a continuous path $q:[0,1]\rightarrow X$ such that $q(0)=A$ and $q(1)=B$? Can you find a continuous path between $f(A)$ and $f(B)$ then?

14. Nov 17, 2015

### BiGyElLoWhAt

Yes, I believe.
I mean, intuitively, yes. The problem lies in the proof part, I think.
Am I supposed to use $f(q): [0,1] \to Y$ and then just the fact that q is continuous and f is continuous therefore f(q) is continuous?

15. Nov 17, 2015

### micromass

What is $f(q)$ supposed to mean? Do you mean $f\circ q$?

16. Nov 17, 2015

### BiGyElLoWhAt

Yes, sorry. The composite.

17. Nov 17, 2015

### BiGyElLoWhAt

I guess it would also be : [q(0), q(1)], would it not?

18. Nov 17, 2015

### micromass

That's an interval, that has no meaning in general topological spaces.

19. Nov 17, 2015

### micromass

Yes. And yes $q$ is continuous (by definition of being a path) and $f$ is continuous (given), so there composition $f\circ q$ is too.

20. Nov 17, 2015

### BiGyElLoWhAt

Is that literally it?