Lebesgue topological dimension

  • Thread starter hofhile
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  • #1
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Hi,

I was reading the definition of dimension from the book: "Topology", Munkres, 2nd ed.
Surely I don't understand, but I wonder how ℝ2 can have dimension 2.

Take the open sets [tex] U_n=\{(x,y)\mid -\infty < x <\infty, n-1<y<n+1\} [/tex] for every integer n. It covers the plane but its order is 2, so the dimension should be less than 2.

Shouldn't be the difinition with balls or "squares"?

Thank you.
 

Answers and Replies

  • #2
mathman
Science Advisor
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Hi,

I was reading the definition of dimension from the book: "Topology", Munkres, 2nd ed.
Surely I don't understand, but I wonder how ℝ2 can have dimension 2.

Take the open sets [tex] U_n=\{(x,y)\mid -\infty < x <\infty, n-1<y<n+1\} [/tex] for every integer n. It covers the plane but its order is 2, so the dimension should be less than 2.

Shouldn't be the difinition with balls or "squares"?

Thank you.

What makes you think the dimension is not 2? What is your understanding of dimension?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
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Hi,

I was reading the definition of dimension from the book: "Topology", Munkres, 2nd ed.
Surely I don't understand, but I wonder how ℝ2 can have dimension 2.

Take the open sets [tex] U_n=\{(x,y)\mid -\infty < x <\infty, n-1<y<n+1\} [/tex] for every integer n. It covers the plane but its order is 2
I presume that Munkres defines the "order" of such a collection of sets. What is that definition and how does it follow that this particular collection has "order 2"?

, so the dimension should be less than 2.
Is there a theorem in Munkres that the dimension of a set is strictly less than the order of a covering collection of such sets?

Shouldn't be the difinition with balls or "squares"?

Thank you.
Definitions like this can be done in many different ways.
 

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