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

[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?

 

[micromass]

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]

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]

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]

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]

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

 

[fresh_42]

Remembers me on reading things like: Let $A_{ij}$ be the transposed matrix of $A$.

My guess is he meant connected by paths: https://www.physicsforums.com/threads/symmetry-and-transitivity.843641/#post-5292127
https://www.physicsforums.com/threads/symmetry-and-transitivity.843641/#post-5292127

 

[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.

 

[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?

 

[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?

 

[micromass]

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

 

[BiGyElLoWhAt]

Yes, sorry. The composite.

 

[BiGyElLoWhAt]

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

 

[micromass]

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]

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]

Is that literally it?

 

[BiGyElLoWhAt]

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

 

[micromass]

Yes

 

[WWGD]

If you are not working on the Reals, you may not have an interval, period.

 

[micromass]

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?

 

[BiGyElLoWhAt]

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?

 

[micromass]

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.

 

[BiGyElLoWhAt]

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]

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?

 

[WWGD]

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?

 

[micromass]

-
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 $\mathbb{R}^n$.

 

[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.