Simple Analysis/topoplogy question

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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.

 

[HallsofIvy]

"Nrp" and "Mrp" are not universal notations! You really should have a definition. I think you mean N_r(p), the neighborhood of p with radius r, \{q| d(p, q)< r\} 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 N_r(p)\subset U. Now, what can you say about U as a subset of M?

 

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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.

 

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Does anyone have any idea?

 

[micromass]

I don't get it. It doesn't seem true I think. Take U=]0,1[ and N=[-1,2] and M=\mathbb{R}, 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...

 

[raw]

I don't get it. It doesn't seem true I think. Take U=]0,1[ and N=[-1,2] and M=\mathbb{R}, 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\subsetN is equivalent to openness in M then N is open in M."

 

[micromass]

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...

 

[raw]

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.