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

[fresh_42]

I think he / she is talking about this:
https://www.physicsforums.com/threads/symmetry-and-transitivity.843641/#post-5292127

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.

 

[BiGyElLoWhAt]

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

 

[BiGyElLoWhAt]

I think he / she is talking about this:
https://www.physicsforums.com/threads/symmetry-and-transitivity.843641/#post-5292127

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]

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]

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"

 

[fresh_42]

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.)

 

[micromass]

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]

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

 

[BiGyElLoWhAt]

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

 

[micromass]

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]

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]

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]

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]

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 into this at all. I think you are heading in the wrong direction there.

 

[Ralph Dratman]

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.

 

[BiGyElLoWhAt]

Then, no. I don't see how n or induction enters into 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]

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?

 

[BiGyElLoWhAt]

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

 

[fresh_42]

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.