Is the proof $diam(A\cup B) \le diam(A) +diam(B)+d(A, B)$ logically perfect? $(X, d) $ be a metric space and $A, B$ be two non empty bounded subsets of $X$.
To show : $$diam(A\cup B) \le diam(A) +diam(B)+d(A, B)$$
Notations:
$diam(A) =Sup\{d(x,y):x,y\in A\}$
$d(A,B) = Inf\{d(x,y): x\in A, y\in B\}$
My attempt:
$(X, d) $ metric space.$ x, y\in X$
$diam(A\cup B) =sup\{d(x, y) : x, y \in A\cup B\}$
$x, y\in A\cup B $ this implies both $x, y \in A $ or $x, y \in B$ or, one of them belongs to $A$ and other belongs to $B$.
We will consider two cases
Case 1: Both $x, y\in A $ ,(similarly both $x,y \in B$) then
\begin{align} d(x, y)&\le\sup\{d(x, y):x, y \in A\}\\
&=diam(A) \end{align}
$\begin{align} diam(A)&\le diam(A) +diam(B) +d(A, B)\end{align}$
Hence,\begin{align} diam(A) +diam(B) +d(A, B)\end{align}
is an upper bound of $\{d(x,y):x,y\in A\cup B\}.$
And hence, \begin{align} diam (A\cup B) &=sup\{d(x,y):x,y\in A\cup B\}\\
&\le diam(A) +diam(B) +d(A, B) \end{align}
And similarly, $x, y\in B \implies diam(A\cup B \le diam(B)\le diam(A) +diam(B) +d(A, B) $
Case 2: $x, y$ doesn't belongs to the same set.(A Or B)
W.L.O.G we asumme
$x\in A \text{ , }y\in B $, then choose two points $a\in A, b \in B. $
\begin{align}
d(x, y) &\le d(x, a) +d(a, b)+d(b, y)\\
&\le diam(A) +diam(B) +d(a, b) \end{align}
Hence, $d(x, y) - diam(A) - diam(B) $ is a lower bound of$ \{d(a,b):a\in A,b\in B \}$.
So, \begin{align}d(x, y)-diam(A) -diam(B) &\le Inf \{d(a,b):a\in 
A,b\in B \} \\ 
&\le d(A, B) 
\end{align}
Hence, $ diam(A) +diam(B) +d(A, B)$ is an upper bound of the set $\{d(x,y):x,y\in A\cup B\}$
And so,\begin{align}diam(A\cup B) &= sup\{d(x,y):x,y\in A\cup B\}\\
&\le diam(A) +diam(B) +d(A, B)\end{align}.
I want to know , is there any logical gaps?
Is my proof correct?
How many marks do you want to give me out of 5 ?
Please verify the proof. Thanks.
 A: There are several problems with your proof.
It begins with “Step 1: $d(x, y) \in \{d(x,y):x,y\in A\cup B\}$”. This is meaningless. What are $x$ and $y$?
This is followed by “Case 1”. At least, there should be a Case 2 after that.
I would perhaps give $3$ marks in $5$, at most.
A: Overall, you should read through your proof and confirm that the following holds:

*

*Every variable is (clearly) defined.  Every variable should be introduced before being used and we should be told in which set it lives.


*Make sure that every line is needed.  There are some statements that are never used or lines that don't follow from the previous ones.  Look forward to identify where each line is used and identify exactly which lines lead to each subsequent line.  Make sure that each conclusion is possible from your assumptions (since you assume $x,y\in A$ for Case 1, you can't conclude anything about points that are not in $A$).


*Make sure that the overall plan is made clear.  You have cases, but the explanation of how those cases fit together to give the proof is missing.
There are several instances of each of these issues in your proof and you should check for them yourself.
