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
Homework Helper
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
Homework Helper
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
$$[a,b] = \{ta+(1-t)b~\vert~t\in [0,1]\}$$