Question

A is not bounded if diam A??​

Answer 1

Answer:

I think you mean diam(S) rather than dim(S), these mean very different things. The usual definition of the diameter of a set in a metric space is

diam(S)=supa,b∈Sd(a,b).

Here the metric under consideration is d(a,b)=|a−b|, the Euclidean metric on R.

To show that the two notions of boundedness are the same, we will show that (1) if diam(S)<∞, then S⊆[−M,M] for some M≥0, and (2) if S⊆[−M,M], then diam(S)<∞.

(1) Take any point x∈S. By definition of the diameter, S⊆[x−diam(S),x+diam(S)]. This, in turn, is contained in the interval [−M,M] if we set M=|x|+diam(S).

(2) If S⊆[−M,M], then diam(S)≤diam([−M,M])=2M<∞.

Step-by-step explanation:

Edit in response to further question: As stated, the question only makes sense for R, because it is not clear what one means by the absolute value of an element in an arbitrary metric space. However, there is a straightforward generalization if one considers a metric space induced by a norm. This consists of a vector space S equipped with a norm function mapping x∈S to a non-negative real number ∥x∥≥0 such that ∥ax∥=|a|∥x∥ for all real numbers a and such that ∥x+y∥≤∥x∥+∥y∥ and ∥x∥=0 implies x=0. Then the metric is defined to be d(x,y)=∥x−y∥. For any metric space induced by a norm in this manner, the same proof as above shows that the diameter of a set is finite if and only if the norm is bounded on that set.

Hope it helps you! ✌️