Don't Understand proof Given (Metric Spaces) everyone I am having trouble understanding this proof:


I don't really understand the proof of Theorem 1.1.  I don't understand how the symmetry axiom for a metric space is just clear by definition.  I also don't really understand how they are proving the triangle inequality axiom for a metric space either (like I don't understand how C is in the r+s neighborhood for A and how A is in the r+s neighborhood for C).  Could someone please explain the proof of these two axioms clearly for me.  Please note that I am an only in an introductory course on analysis so I am still getting used to the abstraction. 
Thanks!
 A: When you are saying that you have not understood the proof to theorem $1.1$, then I assume that you have understood the definition of the Hausdorff metric. 
Then, $d_H(A,B) = \max\left\{\displaystyle\sup_{x \in A} d(x,B),\displaystyle\sup_{y \in B} d(y,A)\right\}$. 
What do we need to verify for $d_H$ to be a metric? We want to show that $d_H$ satisfies the three properties.
What does symmetry mean? It means that $d_H(A,B) = d_H(B,A)$ should be true for all $B , A$. Now, let us literally write down the definitions of these, in terms of what your textbook has given:
$$
d_H(A,B) =\max\left\{\displaystyle\sup_{x \in A} d(x,B),\displaystyle\sup_{y \in B} d(y,A)\right\} \\
d_H(B,A) = \max\left\{\displaystyle\sup_{x \in B} d(x,A),\displaystyle\sup_{y \in A} d(y,B)\right\} 
$$
If we just switch the variables $x$ and $y$ above, we get:
$$
d_H(B,A) = \max\left\{\displaystyle\sup_{y \in B} d(y,A),\displaystyle\sup_{x \in A} d(x,B)\right\}
$$
The difference between $d_H(B,A)$ and $d_H(A,B)$ is just the order in which the maximum is being taken. But then, that doesn't make a difference in our case : $\max\{2,4\} = \max\{4,2\} = 4$, regardless of order. This shows that the above two quantities are equal. This was supposed to be obvious according to the textbook , but I hope you have understood now. If not, then intimate me.
The triangle inequality naturally requires more work in your case, since you say you are only just starting out. We'll do it in detail.
We are supposed to prove that $d(A,C) \leq d(A,B) + d(B,C)$ for any sets $A,B,C$. 
Now, I'll decode the textbook line by line. "If $C$ is in the $r$-neighbourhood of $B$", means that $C \subset \bigcup_{x \in B} B_x(r)$. But what does this mean? This means that for any point $c$, $c$ is in the union of some sets, so $c$ is in at least one of them. That is, there exists $b \in B$ such that $c \in B_b(r)$, which is equivalent to saying that there is some $b \in B$ such that $d(b,c) < r$. In short, this line says:

So for every point $c\in C$, there is some point $b$ such that $d(b,c) < r$. This statement is recorded as $(1)$ for later use.

Similarly, if "$B$ is in the $s$- neighbourhood of $A$", then this means that:

For all $b \in B$, there exists $a \in A$ such that $d(a,b) < s$. this statement is recorded as $(2)$ for later use.

Now we come to the crucial line : "then $C$ is in the $r+s$-neighbourhood of $A$".
So the textbook is saying that the previous two statements imply this : "for every $c \in C$ there exists $a \in A$ such that $d(c,a) < r+s$". Let us call this statement $3$, and we are to show that this is true. Why would this be true? For this, let's use statements $1$ and $2$.
Using statement $1$, we can conclude that for this $c$, there is a $b_c$ (by this, I mean that $b_c$ depends on $c$ : if $c$ changes, so will $b_c$), such that $d(b_c,c) < r$.
Using statement $2$, we can conclude that for $b_c$, there is  an element $a_{b_c}$ (of course, again depending on $b_c$) such that $d(a_{b_c},b_c) < s$. 
Finally, use the triangle inequality to conclude that:
$$
d(a_{b_c},c) \leq d(a_{b_c},b_c) + d(b_c,c) < r + s
$$
Therefore, you see that statement $3$ is true.
I urge you to repeat this argument to go from $A$ to $C$ now. This will show that the Hausdorff distance between these two is less than $r+s$, as desired.
You do not seem to have a problem with $d(A,B) = 0 \iff A =B$. But if you do, then reply back, and I will edit my answer.
EDIT : So it  seems that you want to know  why $d_H(A,B) = 0$ implies that $A=B$. It is obvious that $d_H(A,A) = 0$, since the furthest away that any point in $A$ is from $A$, is precisely zero (how far is it from itself?)
Now, suppose that $d_H(A,B) = 0$. This would mean that $\sup_{x \in A} d(x,B) = 0$, by definition of the Hausdorff metric, which then simply means that $d(a,B) = 0$ for all $a \in A$. But what does this mean? This means by definition of $d(a,B)$ that $\inf_{b \in B} d(a,b) = 0$. Now,  the infimum being zero does not mean that every element is zero. It means, that there is a sequence $b_n$ such that $d(a,b_n)$ decreases to zero. But then, this is the same as saying that $a$ is a limit point of $B$. Therefore, $A$ is a subset of the set of limit points of $B$, hence $A \subset \overline B$, the closure of $B$. 
Similarly, we can conclude that $B \subset \overline A$. But then, $A,B$ are closed sets! So this shows that $$A \subset \overline B  = B\subset \overline A = A$$
Hence, $A=B$ , as desired.  
$()$
A: *

*Symmetry:
From the definition of the Hausdorff metric, in (2) or (4) you should be able to see that swapping the argument order (i.e swapping A and B) does not change the value assigned.

*Triangle inequality:
This comes from the triangle inequality on the base metric space. To show that $d_h(A,B) + d_h(B,C) \geq d_h(A,C)$, note that every element $c \in C$ is within distance $r = d_h(B,C)$ of some element $b \in B$, which in turn is within distance $s = d_h(A,B)$ of some element $a \in A$. By the triangle inequality of the base metric on $X$, we have: $$d(c,A) \leq d(c,a) \leq d(c,b) + d(b,a) \leq d_h(B,C) + d_h(A,B) = r+s$$ 
Because this is true for all $c \in C$, it is also true for the supremum over all c. Swapping A and C gives the same result, so both $\sup_{c\in C}(c,A)$ and $\sup_{a\in A}(a,C)$ are less than $ r+s$, therefore so is $d_h(A,C)$ (from definition (4) in the notes). 
