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

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?

Sorry, I'm not understanding this at all.

Gold Member
The idea was to show continuity between a 2 points in X after they have been mapped to Y. So wouldn't that be my interval?
##q : [A,B] \to f\circ q : [q(A),q(B)] ##
Or something, I'm not really good with the rigorous math notation. I hope you can interpret this as I think it would be.

WWGD
Gold Member
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?
Sorry, I was referring to OP, trying to get him/her to clarify the assumptions.

EDIT: Besides, how do we know s/he is working in a topological vs?

Last edited:
WWGD
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?
I don't know what you mean by continuity between two points. Would you clarify?

-
EDIT: Besides, how do we know s/he is working in a topological vs?

In post 12, the OP specified ##X## and ##Y## to be subspaces of ##\mathbb{R}^n##.

• BiGyElLoWhAt
fresh_42
Mentor
• BiGyElLoWhAt
Gold Member
By continuity I mean a continuous path betwwen points, so a continuous curve, or map (maybe I want to use the word transform here).

Gold Member

I'm not certain about the correct English wording, we call it path connected.
So A ~ B iff there is a path from A to B. As far as I understood it it's to show that f(A) ~ f(B) for continuous f. His / her intervals are the parametrization of the paths.
Exactly. Thank you for clarifying. You can use he, also, by the way.

WWGD
Gold Member
Ah, I see, sorry. So you are asking whether continuity preserves path-connectedness? If that is the question, then the answer is then no; the topologist's sine curve is a counterexample.

FactChecker
Gold Member
If you are using the ε, δ definition of continuous functions, you will have to trace the ε, δs step by step through the composition of the functions. If you are using the open set definition (the preimage of every open set is open), then the proof is easier. Just say that for every open set, O, in the range of f(g), f-1(O) is open because f is continuous; and g-1(f-1(O)) is open because g is continuous, so f(g) is continuous.

The only thing remaining is to verify the end-point condition of the definition of "homologous"

Last edited:
• BiGyElLoWhAt and fresh_42
fresh_42
Mentor
A ~ B ⇒ f(A) ~ f(B) for f continuous

... so f(g) is continuous.
... So taking g as the path between A and B, f⋅g defines the path between f(A) and f(B).

(In the ε-δ-world it's probably faster to show the equivalence to the open set definition rather than stepping through.)

Last edited:
Ah, I see, sorry. So you are asking whether continuity preserves path-connectedness? If that is the question, then the answer is then no; the topologist's sine curve is a counterexample.

But continuity does preserve path connectedness

WWGD
Gold Member
Yes, you're right, I did something wrong somewhere, let me double-check.

Gold Member
So out of curiosity, would the epsilon delta prrof that I provided work?

So out of curiosity, would the epsilon delta prrof that I provided work?

I didn't see any proof from you yet... And I gave you the proof, it's just composition of functions.

fresh_42
Mentor
So out of curiosity, would the epsilon delta prrof that I provided work?
I don't know what you mean?

For the ε-δ-definition of continuity you need a metric space. Then both definitions are almost obviously equivalent. E.g. the neighborhood { x with | x - z| < δ } of z defines an open set. And in each open set with z in it you can find a small enough neighborhood of z.

On the other hand continuity is defined for all topological spaces (via the open set definition: ##f^{-1} (N)## is open for all open ##N##).

fresh_42
Mentor
I see. You meant your first post. It's far from being precise enough. Essentially you can cobble your path with overlapping neighborhoods but only saying by induction wouldn't be enough to me. But that's my opinion

Btw. you messed your path up, too:

The idea was to show continuity between a 2 points in X after they have been mapped to Y. So wouldn't that be my interval?
##q : [A,B] \to f\circ q : [q(A),q(B)] ##
Or something, I'm not really good with the rigorous math notation. I hope you can interpret this as I think it would be.

It has to be ##q : [0,1] \to [A,B] ⇒ f\circ q : [0,1] → f ([q(0),q(1)]) = [f(A),f(B)] ##. And ##f## has to be continuous for ##f\circ g## being continuous.

• BiGyElLoWhAt
Gold Member
Btw. you messed your path up, too:
Ahh, yea. I see it now. I think I was mixing up what each term represented. I think I see what you mean. I'll post back with an attempt at the full proof for the epsilon delta idea. I just want to do it as an exercise at this point.

FactChecker
Gold Member
So out of curiosity, would the epsilon delta prrof that I provided work?
If the proof you are referring to is
if f is continuous, then there exists an ϵ for every δ such that ... blah...so
f(A)∼f(A+ϵ)∼f(A+nϵ) and by induction f(A)∼f(B) for large enough n
Then, no. I don't see how n or induction enters in to this at all. I think you are heading in the wrong direction there.

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.
Of course it really should not matter which letter is used to stand for which quantity. Maybe the test-writer wanted to see if students could see their way past the specific letters to the concepts involved.

Gold Member
Then, no. I don't see how n or induction enters in to this at all. I think you are heading in the wrong direction there.

So by the definition of continuity, for any ##\delta## there exists an ##\epsilon## such that ##f(x+\epsilon)\leq f(x)+\delta##.

If we follow this logic, we can say that there also exists an ##\alpha## such that ##f((x+\epsilon)+\alpha)\leq f(x+\epsilon)+\delta## for any ##\delta##, and so forth. So, we can just keep adding terms, and eventually we will traverse from ##f(A)\to f(B) ## continuously, so those 2 points must be continuous, if we can trace a path between them.

I understand, now, that this isn't what the course is looking for, but does this not work as well?

WWGD
Gold Member
So by the definition of continuity, for any ##\delta## there exists an ##\epsilon## such that ##f(x+\epsilon)\leq f(x)+\delta##.

If we follow this logic, we can say that there also exists an ##\alpha## such that ##f((x+\epsilon)+\alpha)\leq f(x+\epsilon)+\delta## for any ##\delta##, and so forth. So, we can just keep adding terms, and eventually we will traverse from ##f(A)\to f(B) ## continuously, so those 2 points must be continuous, if we can trace a path between them.

I understand, now, that this isn't what the course is looking for, but does this not work as well?

You are using an expression " those 2 points must be continuous" I never heard of. would you explain what you mean by it?

Gold Member
There must be a continuous path between the two points, meaning they are connected, sorry.

fresh_42
Mentor
There must be a continuous path between the two points, meaning they are connected, sorry.
You should recap the definitions and concepts (in this order) of continuity (metric and topological), connection and path connection.
To construct a path from ##f(A)## to ##f(B)## using the ε-δ-definition of continuity you have to be more careful and must assure that your path doesn't run out of the domain and that it really leads from ##f(A)## to ##f(B)##. Construct it piece wise by transferring a sufficiently number of steps from ##A## to ##B## into your target path. Just saying "enough δ's will get us enough ε's" which you basically did isn't enough. E.g. where have the absolute values gone that the ε-δ-definition of continuity consists of?
I recommend you to draw some figures of all and decide on them what to do.

• BiGyElLoWhAt