- #1
Assume that f is a continuous, real-valued function
- Thread starter Ric-Veda
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- Continuous Function
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Homework Help Overview
The discussion revolves around the continuity of a real-valued function defined on a metric space and the behavior of sequences converging to a point within that space. Participants are tasked with proving that the limit of the function values at the sequence converges to the function value at the limit point.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking
Approaches and Questions Raised
- Participants discuss the necessity of using definitions related to continuity and limits, questioning the selection of epsilon and the implications of the proof strategy. There are suggestions to follow a structured proof approach and considerations about the applicability of the proof in different contexts.
Discussion Status
The discussion is ongoing, with participants providing various attempts at the proof and questioning each other's reasoning. Some guidance has been offered regarding the structure of the proof, but there is no explicit consensus on the correctness of any particular approach yet.
Contextual Notes
Participants note the importance of including quantifiers in their arguments and the implications of continuity in the context of metric spaces. There is also mention of the need for clarity in the proof regarding the assumptions made.
- #2
Svein
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The usual way is to use the definitions. What you need to prove is that given any ε>0, there is an N such that (\forall n>N)(\vert f(x_{n})-f(x)\vert <\epsilon). You do this in steps:
- First, since f is continuous, you can find a δ...
- Then since \lim_{n\rightarrow\infty}x_{n}=x, there is an N such that...
- Therefore...
- #3
member 587159
Ric-Veda said:
Anu proof will have to use the definition. Where did you pick an epsilon greater than 0? It is weird to pick ##n##. I suggest you follow the proof strategy suggested by @Svein, and after you are done with the proof, convince yourself that the same proof also applies where the codomain is an arbitrary metric space (i.e. change the absolute values with a distance function and look careful that you never used something else)
- #4
- #5
member 587159
Ricster55 said:
Why ##\exists n_0: |f(x_n)-f(x)| < \epsilon \forall n \geq n_0##? That shouldn't be included in your proof at the beginning, unless you explicitely say that this is what you are going to prove. Also, your lack of the ##\forall x \in X## quantifier is not so good. You should include it. But for the rest, the argument seems fine.
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