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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.
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 was referring to OP, trying to get him/her to clarify the assumptions.True, but in a vector space you usually have the following notation
[tex][a,b] = \{ta+(1t)b~\vert~t\in [0,1]\}[/tex]
so perhaps he meant that?
I don't know what you mean by continuity between two points. Would you clarify?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?

EDIT: Besides, how do we know s/he is working in a topological vs?
Exactly. Thank you for clarifying. You can use he, also, by the way.I think he / she is talking about this:
https://www.physicsforums.com/threads/symmetryandtransitivity.843641/#post5292127
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.
... So taking g as the path between A and B, f⋅g defines the path between f(A) and f(B).... so f(g) is continuous.
Ah, I see, sorry. So you are asking whether continuity preserves pathconnectedness? If that is the question, then the answer is then no; the topologist's sine curve is a counterexample.
So out of curiosity, would the epsilon delta prrof that I provided work?
I don't know what you mean?So out of curiosity, would the epsilon delta prrof that I provided work?
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.
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.Btw. you messed your path up, too:
If the proof you are referring to isSo out of curiosity, would the epsilon delta prrof that I provided work?
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.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
Of course it really should not matter which letter is used to stand for which quantity. Maybe the testwriter wanted to see if students could see their way past the specific letters to the concepts involved.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.
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?
You should recap the definitions and concepts (in this order) of continuity (metric and topological), connection and path connection.There must be a continuous path between the two points, meaning they are connected, sorry.