Prove that if a subset $A$ of a metric space is bounded then the closure of $A$ is bounded and the diameter of $A$ is equal to the diameter of the closure of $A$.
This is the question I am working on at the moment. I think I have proven that cl(A) is bounded as follows:
If A is bounded then $\exists x_0 \in X$ and $K$ constant s.t. $d(x_0,x)\leq K$ $\forall x\in X$.
Let $a \in$ cl(A), then $B_\epsilon (a) \cap A \neq \emptyset$ $\forall \epsilon > 0$.
Now let $a'\in B_\epsilon (a) \cap A$, then $d(a,x_0)\leq d(a,a')+d(a',x_0)\leq \epsilon +K$.
As this is true $\forall \epsilon > 0$ this means $d(a, x_0) \leq K$ and cl(A) is hence bounded.
So now I need to show that diam(A) = diam(cl(A)).
The diameter is defined as diam(A) = sup($d(x,y)$) $\forall x,y \in A$.
I want to do this by letting $K$ be the least upper bound of A and then I would like to just say that as both A and cl(A) are bounded by the same thing, the supremum of the distance between any two points in either A or cl(A), which is $K$, is equal to diam(A) and diam(cl(A)). Is this the correct or right way to go about it?