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- Thread starter HeraclitusJ
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This depends on the Heine-Borel theorem, but does it satisfy you? If a subset of R^{n} is bounded, take its closure to get a compact set, and by compactness find a finite covering by ε-balls for any ε > 0.

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Let X be a subset of Rn and assume that it's not totally bounded. Then you can construct an infinite set of points x_n such that d(x_i, x_j)>e for some e>0. The geometric intuition is that if you have infinite number of these points, an infinite number of them must fall on some straight line (because of the finite dimensionality) and then you can use that to show that the diameter of X is not finite thereby completing the proof. The key step that I can't show is that there are an infinity of these points on one straight line.

I'm not sure if this works, but I'd rather not invoke Heine-Borel...

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I have an exam on Wed and I was praticing couple of things. Here is my question,

Suppose X € R^n is compact. Prove that X is bounded.

Lets suppose X is compact.

Let α = supX.

Since X is nonempty for any ε > 0.

There is y € X such that α – ε < y < α

Thus α is a limit point of X which contain all limit points. So α € X.

That proves that X is bounded.

does it make sense? did i do it right?

- #7

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Your proof is invalid. By saying "let α = sup X", there are two things wrong there:

(1) You're already assuming that X is bounded above, and you're trying to prove that X is bounded.

(2) sup X (and bounded above) only makes sense if X is a subset of [itex]\mathbb{R}[/itex] (or some other ordered set with the least upper bound property)!

Additionally, limit points have little to do with boundedness. (What does "α is a limit point of X which contain all limit points" even mean?)

Remember the definition of compactness. The simplest way to solve that problem is to use the definition of compactness directly; the idea is to find an open cover of X such that each set in the cover is bounded. By compactness, finitely many of these sets cover X, and you have to argue that the union of finitely many bounded sets is bounded.

As a specific example, I denote [itex]B(x; 1)[/itex] to mean the ball of radius [itex]1[/itex] centered at [itex]x[/itex] (that is; [itex]B(x; 1) = \{y \in \mathbb{R}^n \mid \lvert x - y \rvert < 1[/itex]); these balls are open sets (by the definition of the topology induced by a metric space). My open cover will be [itex]\mathcal{O} = \{B(x; 1) \mid x \in X\}[/itex]; it's easy to verify that [itex]\mathcal{O}[/itex] covers [itex]X[/itex]. By compactness, there is a finite subcover [itex]\mathcal{F} \subseteq \mathcal{O}[/itex]. Since I picked each set in [itex]\mathcal{O}[/itex] to be bounded, [itex]X \subseteq \bigcup \mathcal{F}[/itex] is bounded.

(1) You're already assuming that X is bounded above, and you're trying to prove that X is bounded.

(2) sup X (and bounded above) only makes sense if X is a subset of [itex]\mathbb{R}[/itex] (or some other ordered set with the least upper bound property)!

Additionally, limit points have little to do with boundedness. (What does "α is a limit point of X which contain all limit points" even mean?)

Remember the definition of compactness. The simplest way to solve that problem is to use the definition of compactness directly; the idea is to find an open cover of X such that each set in the cover is bounded. By compactness, finitely many of these sets cover X, and you have to argue that the union of finitely many bounded sets is bounded.

As a specific example, I denote [itex]B(x; 1)[/itex] to mean the ball of radius [itex]1[/itex] centered at [itex]x[/itex] (that is; [itex]B(x; 1) = \{y \in \mathbb{R}^n \mid \lvert x - y \rvert < 1[/itex]); these balls are open sets (by the definition of the topology induced by a metric space). My open cover will be [itex]\mathcal{O} = \{B(x; 1) \mid x \in X\}[/itex]; it's easy to verify that [itex]\mathcal{O}[/itex] covers [itex]X[/itex]. By compactness, there is a finite subcover [itex]\mathcal{F} \subseteq \mathcal{O}[/itex]. Since I picked each set in [itex]\mathcal{O}[/itex] to be bounded, [itex]X \subseteq \bigcup \mathcal{F}[/itex] is bounded.

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wow its wrong? huh what world am i living in?

ps: thats alpha.Let α = supX.

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