Simple Analysis/topoplogy question

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Homework Help Overview

The discussion revolves around a proof involving metric spaces, specifically addressing the openness of a metric subspace N within a larger metric space M. The original poster seeks to establish the conditions under which N is considered open in M given that a set U is open in both N and M.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the definitions of open sets in metric spaces and question the clarity of the original proposition. There is discussion about the notation used for neighborhoods in the context of the problem. Some participants express confusion regarding the validity of the proposition and provide counterexamples to challenge its truth.

Discussion Status

The discussion is ongoing, with some participants providing insights and clarifications about the definitions involved. There is a recognition of the potential misunderstanding of the original question, and some participants seem to arrive at a clearer interpretation of the problem, although no consensus has been reached yet.

Contextual Notes

There are concerns about the notation used for neighborhoods and the implications of the openness of sets in different metric spaces. Some participants question the assumptions underlying the original statement and explore the implications of specific examples that may contradict the proposition.

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Homework Statement


Let N be a metric subspace of M. Prove that if U is open in N and open in M, then N is open in M.


Homework Equations


U is open in N if for every p in U there exists an r>0 such that Nrp is a subset of U. Similarly, r is open in M if for every p in U there exists an r>0 such that Mrp is a subset of U.



The Attempt at a Solution


I wrote the definitions down and know that Nrp=Mrp but don't know where to go beyond that.
 
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"Nrp" and "Mrp" are not universal notations! You really should have a definition. I think you mean [itex]N_r(p)[/itex], the neighborhood of p with radius r, [itex]\{q| d(p, q)< r\}[/itex] but that notation would be the same in either space M or space N.

Also, you statement of the proposition, "Prove that if U is open in N and open in M, then N is open in M" makes no sense. Do you mean "if U is open in N, then it is open in M?

Let U be open in N. Let p be a member of U. Then there exist r> 0 such that [itex]N_r(p)\subset U[/itex]. Now, what can you say about U as a subset of M?
 
You are right about the notations, I didn't know how to put subscripts in. No, I already proved that if U is open in N then it is open in M. I need to prove that if U is open in N and in M and N is a metric subspace of M then N is open in M.
 
Does anyone have any idea?
 
I don't get it. It doesn't seem true I think. Take U=]0,1[ and N=[-1,2] and M=[tex]\mathbb{R}[/tex], then U is open in N and M, but N is not open in M. Or do I fail to understand the question properly??

Or do you mean: if every open set of N is open in M, then N is open in M? That seems trivial however...
 
micromass said:
I don't get it. It doesn't seem true I think. Take U=]0,1[ and N=[-1,2] and M=[tex]\mathbb{R}[/tex], then U is open in N and M, but N is not open in M. Or do I fail to understand the question properly??

Or do you mean: if every open set of N is open in M, then N is open in M? That seems trivial however...

Yes, I think that's what the question is asking. It's a weird question but my professor said it should be simple yet I am not really getting anywhere.

The questions reads:
"Prove that if openness of S[tex]\subset[/tex]N is equivalent to openness in M then N is open in M."
 
Well if it's the question: If every open set in N is open in M, then N is open in M.

The answer is not so hard. N is open in N. So the condition says that N is open in M. Which we needed to show...
 
micromass said:
Well if it's the question: If every open set in N is open in M, then N is open in M.

The answer is not so hard. N is open in N. So the condition says that N is open in M. Which we needed to show...

Oh wow, I'm smacking myself in the head right now, so simple. Thank you very much for taking the time to answer my question.
 

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