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

Gold Member
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?

micromass
Staff Emeritus
Homework Helper
You will have to say what ##X## and ##Y## are, and what ##\sim## is.

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

micromass
Staff Emeritus
Homework Helper
At first I thought I was going to make an effort to understand the OP, but I just gave up.

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.

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

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

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

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

micromass
Staff Emeritus
Homework Helper
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.

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

S.G. Janssens
You mean the following?
Yes, precisely.

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

S.G. Janssens
Gold Member
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.

micromass
Staff Emeritus
Homework Helper
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?

Gold Member
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?

micromass
Staff Emeritus
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What is ##f(q)## supposed to mean? Do you mean ##f\circ q##?

Gold Member
Yes, sorry. The composite.

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

micromass
Staff Emeritus
Homework Helper
I guess it would also be : [q(0), q(1)], would it not?

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

micromass
Staff Emeritus
Homework Helper
Yes, sorry. The composite.

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.

BiGyElLoWhAt
Gold Member
Is that literally it?

Gold Member
Do I not need to specify my interval that I'm mapping over to show continuity between two points?

micromass
Staff Emeritus
Homework Helper
Yes

BiGyElLoWhAt
WWGD
Gold Member
If you are not working on the Reals, you may not have an interval, period.

micromass
Staff Emeritus
Homework Helper
If you are not working on the Reals, you may not have an interval, period.

True, but in a vector space you usually have the following notation
$$[a,b] = \{ta+(1-t)b~\vert~t\in [0,1]\}$$
so perhaps he meant that?

Gold Member
Well if I'm looking between two points, and then looking between a map of those two points, wouldn't those be my 2 respective intervals?