Totally bounded but not bounded

[johnqwertyful]

It seems strange, but would a metric space consisting of two points, X={a,∞} be totally bounded, but not bounded? because d(a,∞)=∞. But for all ε>0, X=B(ε,a)UB(ε,∞).

It's been proven that totally bounded→bounded, so this is wrong. Why?

 

[johnqwertyful]

http://www.math.ucla.edu/~tao/resource/general/121.1.00s/compact.pdf

in this proof of totally bounded implies bounded, he takes the maximum of d(xi,x1), which he assumes to be finite. So if we assume that the distance between every two points is finite, then we get totally bounded implies bounded.

 

[jgens]

So if we assume that the distance between every two points is finite

This "assumption" is incorporated into the definition of metrics.

 

[shortydeb]

the axioms for a metric space state that for any two points in the metric space, their distance is a real (and finite) number.

 

[blue_raver22]

While it may seem counterintuitive, a metric space consisting of two points, X={a,∞}, is indeed both totally bounded and bounded. This is because the definition of a totally bounded space is one in which every ε-ball contains a finite number of points, not necessarily a finite distance. In this case, the ε-ball around the point ∞ would contain only one point, while the ε-ball around the point a would contain both points a and ∞. Therefore, the space is totally bounded.

Additionally, the statement that "totally bounded→bounded" is incorrect. The correct statement is "bounded→totally bounded". This means that a bounded space is also totally bounded, but a totally bounded space may not necessarily be bounded. In other words, being bounded is a stronger condition than being totally bounded.

In summary, the metric space X={a,∞} is both totally bounded and bounded, despite the distance between the two points being infinite. This is because the definition of totally bounded only concerns the number of points within an ε-ball, not the distance between them.