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 $\mathbb{R}^n$.