example of a totally bounded but not bounded

Question

I have got an question:-

is there any counter-example of totally bounded but not bounded space

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can anyone help me please.thanks for your help

Answer 1

No there is no counterexample, total bounded implies boundness, when there is a finite cover using $\epsilon$ neighborhoods, there is a cover using only a single neighborhood which is still bounded.

Answer 2

SEE http://en.wikipedia.org/wiki/Totally_bounded_space#Definition_for_a_metric_space

Totally bounded implies bounded. These are terms used in the setting of a metric space. So there is no counterexample such as you request.

I will show by induction that the union of finitely many bounded sets is still bounded. Bounded means bounded diameter, any two points of the set are no farther apart than some $D.$ The least upper bound among the $D$ that work for the set is called the diameter of the set.

Suppose we have two nonempty sets in a metric space, $A$ has diameter $D_A,$ set $B$ has diameter $D_B.$ Furthermore, there are a point $a \in A$ and $b \in B.$ There is some distance between them, $$ d(a,b) = W. $$ As a result, any two points in the union are no farther apart than $$ W + D_A + D_B $$ which is an upper bound for the diameter of the union. So the union has a diameter also.

The induction step is to add in a third set, get a bigger diameter , then a fourth, and so on. As long as the number of sets in the union is finite, the result has a finite diameter, i. e. is bounded.

The definition of totally bounded is that for any $\epsilon > 0,$ the space is covered by finitely many $\epsilon$-balls. But, by definition and the triangle inequality, each $\epsilon$-ball has diameter no larger than $2 \epsilon.$ Since finitely many of these cover the entire metric space, it also is bounded, i.e. has a finite diameter.