# Proof verification - diam$(Y) =$ diam$(\bar{Y})$

The identity I was asked to prove is as follows:
Let $$(X, d)$$ be a metric space and $$Y \subset{X}$$. Prove that $$diam(Y) = diam(\bar{Y})$$ where $$diam(Y) = sup\{ d(x,y)\ |\ x,y \in Y \}$$
My proof is this:

$$Y \subset \bar{Y} \implies diam(Y) \le diam(\bar{Y})$$
Therefore, to prove $$diam(Y) = diam(\bar{Y})$$ it suffices to show that assuming $$diam(Y) \lt diam(\bar{Y})$$ leads to contradiction.

Suppose $$diam(Y) \lt diam(\bar{Y})$$. Then $$\exists\ a \in \bar{Y}$$ such that $$B(a,r)\ \cap\ Y = \emptyset$$ for some $$r \gt 0$$. However, $$\bar{Y}$$ consists of all points adherent to Y, or
$$\bar{Y} = \{x \in X\ |\ B(x,r)\ \cap\ Y \neq \emptyset, \forall\ r \gt 0 \}$$

Thus, $$\bar{Y}$$ must contain a point which is not adherent to $$Y$$, which is a contradiction of its definition.
Thus, $$diam(Y) = diam(\bar{Y})$$ $$\blacksquare$$

Is this proof correct? or are there faults\inconsistencies in it?

• How do you prove the assumption "there exists $a \in \bar{Y}$ and $r>0$ such that $B(a,r)\cap Y = \varnothing$"? Jan 24, 2022 at 21:00

I'd they the general proof is correct. The only step you could make a bit more precise is

Suppose $$\mathrm{diam}(Y)<\mathrm{diam}(\bar Y)$$. Then $$∃ a∈\bar Y$$ such that $$B(a,r) ∩ Y=∅$$ for some $$r>0$$.

To me, it is not immediately clear why this is true (it is -- but it requires some intermediate steps (at least for me)).

If you want to stick closer to the definitions here, I would probably start by saying that $$\mathrm{diam}(Y)<\mathrm{diam}(\bar Y)$$ implies that there exist some $$x,y \in \bar Y$$ with $$d(x,y) > \mathrm{diam}(Y)$$ and then deduce a contradiction from there.

By definition $$\bar{Y}$$ contains elements of $$Y$$ or elements of the complement of Y , $$Y^c$$, such that $$y\in Y^c$$ and $$B(y,r) \cap Y \neq \emptyset \forall r>0$$.

Suppose $$a\in Y$$ and $$b \in Y_c \cap \bar{Y}$$. Then $$inf \ d(a,b)=0$$.

As mentioned above, $$diam(Y)\le diam(\bar{Y})$$.

$$\bar{Y}$$ is closed, so the supremum of the distance between elements is achieved by elements of that set. So select $$p,q \in \bar{Y}, d(p,q)=diam(\bar{Y})$$.

If $$p,q \in Y$$, we are done.

Suppose $$p,q\in Y^c\cap \bar{Y}$$ where $$d(p,q)=diam(\bar{Y})$$. Such points exist because $$\bar{Y}$$ is closed.

$$diam(\bar{Y})\le diam(Y)$$ as mentioned.

Then by definition:

$$d(p,q)\ge sup\ d(a,b), a,b\in Y$$

$$d(p,q)\le d(p,a)+d(a,b)+d(b,q)$$ by the triangle inequality.

Suppose $$d(b,q)$$=r. We are guaranteed a $$b_2\in Y$$ so that $$d(b_2,q)\le r/2$$. This process can be iterated, therefore, if $$r>0$$, we do not have a least upper bound to the right side of the inequality. By the same reasoning, we can get a point to substitute for $$a$$. Similar reasoning applies if either $$p$$ or $$q$$ is in $$Y$$ because then $$d(a,p)=0$$.

Take the supremum of both sides: then $$sup \ d(p,q)\le sup \ d(a,b)$$.

If less, this contradicts that $$Y$$ is a subset of $$\bar{Y}$$. It follows that the supremema and therefore the diameters are equal.