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?

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

2 Answers 2


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.


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