Undergrad How Do Metric Space Properties Relate to Continuity?

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The discussion focuses on the properties of metric spaces, specifically relating compactness, completeness, connectedness, and separability to the concept of continuity in the real line. It clarifies that a connected subset of the real line is an interval, while a compact set is characterized as closed and bounded according to the Heine-Borel theorem. The conversation also addresses the separability of subsets of the reals and confirms that every subset is separable and complete if closed. Additionally, it explores the metric space properties of the complex numbers, concluding that the complex plane is separable and complete. Overall, the thread emphasizes the application of metric space properties in understanding continuity and related concepts.
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Hello. The questions i make here in this thread are basically about like explanations of topics on metric spaces. We know about compactness, completeness, connectedness, separatedness, total boundedness of metric spaces. I know that continuity of the real line means that it has no gaps. What could we say about the properties of the metric spaces i described above in the spirit of the description of the continuity of the real line? I am not talking about the definition which is an abstraction, i am talking about the application of the definition like above in the real line. Thank you.
 
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The connected and compact subset of the real numbers can be characterised as follows:

Connected subset of the real line = interval
Compact set = closed + bounded subset (Heine Borel theorem).

What is your definition of "separated metric space" or what do you mean when talking about separatedness?
 
Math_QED said:
The connected and compact subset of the real numbers can be characterised as follows:

Connected subset of the real line = interval
Compact set = closed + bounded subset (Heine Borel theorem).

What is your definition of "separated metric space" or what do you mean when talking about separatedness?
Thank you. I am sorry i made mistakes. The correct are separable and separability not what i wrote. What about completeness( if i remember correctly it is about sequences, cauchy sequences, convergence)
 
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universe function said:
Thank you. I am sorry i made mistakes. The correct are separable and separability not what i wrote. What about completeness( if i remember correctly it is about sequences, cauchy sequences, convergence)

Every subset of the reals is separable. A subset of the reals is complete if and only it is closed.
 
I have another question but is a little off topic I think. Is C which is the set of complex numbers equipped with the metric that is related to the norm, d(x,y)=llx-yll2=√((x1-x0)2+(y1-y2)2), where x=(x1,x2), y=(y1,y2) a metric space? Is it separable? Is it complete if and only if it is closed? Excuse me if these questions have as answer no.
 
universe function said:
I have another question but is a little off topic I think. Is C which is the set of complex numbers equipped with the metric that is related to the norm, d(x,y)=llx-yll2=√((x1-x0)2+(y1-y2)2), where x=(x1,x2), y=(y1,y2) a metric space? Is it separable? Is it complete if and only if it is closed? Excuse me if these questions have as answer no.

Yes, it is a metric space. It is separable. Can you think of a countable dense subset? Hint: Use density of ##\Bbb{Q}## in ##\Bbb{R}##. It is definitely complete, because ##\mathbb{R}## is complete. Asking that it is closed makes little sense because every topological space is automatically closed in itself.
 
Math_QED said:
Yes, it is a metric space. It is separable. Can you think of a countable dense subset? Hint: Use density of ##\Bbb{Q}## in ##\Bbb{R}##. It is definitely complete, because ##\mathbb{R}## is complete. Asking that it is closed makes little sense because every topological space is automatically closed in itself.
Thank you, perhaps the answer is that ℚ2 is dense in ℝ2? and because ℝ2 is homeomorphic to ℂ then ℚ2 is dense in ℂ?
 
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universe function said:
Thank you, perhaps the answer is that ℚ2 is dense in ℝ2? and because ℝ2 is homeomorphic to ℂ then ℚ2 is dense in ℂ?

Yes, that's exactly it!
 
Math_QED said:
Yes, that's exactly it!
Oh, i answered it correctly.
 
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