How do you define unboundedness in Euclidean space?

  • #1
Eclair_de_XII
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TL;DR Summary
On the real numbers, a set ##D## is bounded iff there exists a positive ##M## such that for all ##y\in D##, ##|y|\leq M##.

On the real numbers, a set ##D## is unbounded iff for all positive ##M##, there is a ##y\in## such that ##|y|>M##.
I read in my textbook Calculus on Manifolds by Spivak that a set ##A\subset \mathbb{R}^n## is bounded iff there is a closed n-rectangle ##D## such that ##A\subset D##. It should be plain that if I wanted to define unboundedness, I should just say something along the lines of: "A set ##A\subset \mathbb{R}^n##is unbounded iff for all n-rectangle ##D##, ##A## is not contained in ##D##.D

So my question is if the following definition of unboundedness is valid: "A set ##A## is unbounded if for all n-rectangles ##D##, there is a ##y\in A## such that there is an open set containing ##y## that is a subset of ##\mathbb{R}^n \cap D^c##".
 
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  • #2
Boundedness is a metric concept, open sets is a topological concept so you shouldn't try to define boundedness via topological properties.

Why don't you take the negation of the definition of boundedness Spivak provides?

Thus, a set ##A## is unbounded if it is not contained in a rectangle.

I.e., for every rectangle ##D## we have that ##A \not\subseteq D##, or equivalently for every rectangle ##D##, there is ##a \in A \setminus D##.

Another fun characteristation of boundedness in this context:

A set ##A## is unbounded if this set contains a sequence of elements ##(a_n)_n## such that ##\Vert a_n \Vert \to \infty##.

On an unrelated note: what is your background on calculus/real analysis? This book my Spivak is hard to read if you don't have the proper background.
 
  • #3
Math_QED said:
Why don't you take the negation of the definition of boundedness Spivak provides?

Thus, a set ##A## is unbounded if it is not contained in a rectangle.

I.e., for every rectangle ##D## we have that##A \not\subseteq D##, or equivalently for every rectangle ##D##, there is ##a \in A \setminus D##.

I guess that would be the most practical option in order to do problems within the book.

Math_QED said:
what is your background on calculus/real analysis?

I took the entire calculus series, an introductory real analysis course, and a vector calculus analysis course. I was told that the latter was the analysis analogue to Calculus III-IV, like how the introductory analysis course was an advanced course elaborating on stuff from Calculus I. The vector analysis course used the book in question, so it's not my first time reading it.

I didn't take the non-introductory analysis course nor the topology course offered at my university during my college career. So I naturally would not have the background necessary to understand why, as you said in the first sentence of your post, I should not attempt to define boundedness with topological properties.
 
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  • #4
"So my question is if the following definition of unboundedness is valid: "A set A is unbounded if for all n-rectangles D, there is a y∈A such that there is an open set containing y that is a subset of Rn∩Dc"."

the answer to your question is "yes".
 
  • #5
Math_QED said:
Boundedness is a metric concept, open sets is a topological concept so you shouldn't try to define boundedness via topological properties.

This is only partially true. If ##(X, \tau)## is a topological vector space, then one naturally defines a set ##A \subseteq X## to be bounded if for every neighborhood ##V## of ##0## there exists a scalar ##\lambda## such that ##A \subseteq \lambda V##. If ##\tau## derives from a norm, this reduces to the familiar definition, but in the general (non-metric) case this definition leads to some interesting properties and the notion of bornology.

(In view of the above, I like Spivak's definition since the general definition is already in there.)
 

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