$\{y \in M : d(x, y) + d(y, z) < a\}$ is open

Question

Let $(M, d)$ be a metric space, $x, z \in M$ and $a > 0$. Prove that the set $A = \{y \in M : d(x, y) + d(y, z) < a\}$ is open.

I'm having a fair bit of trouble trying to find how to use triangle inequality here. Specifically, if I let $r = a - d(x, y) - d(y, z)$ I can't cancel enough terms to get $d(q, y) < a$. For example \begin{align*} d(q, y) \leq d(x, q) + d(x, y) \leq d(x, y) + d(y, q) + d(x, z) + d(z, y) < a + d(x, z) \end{align*} What am I missing, or is my approach just wrong? I've considered letting $r = a - \min\{d(y, x), d(y,z)\}$ and I believe that this works as far as the inequality is concerned but I'm not sure that it actually proves the statement.

Answer 1

Knowing that any metric is continuous with respect to its own topology, in particular, the functions $f(y) = d(x, y) $ and $g(y) = d(z, y)$ are continuous, and then so is their sum $h(y) = f(y) + g(y)$. Hence, $A = h^{-1}\left(-1, a\right)$ is open, being the inverse image of an open set (note that the $-1$ was an arbitrary choice, we only need a negative number at the left end of the interval to include $0$ in our interval).

Answer 2

Simple We need to prove that $A$ is open. That is we need to prove $\forall y \in A, \exists r>0$ such that $B(y,r) \subset A$

So given $y$ we need to find $r$ such that $d(y,q)<r \implies q \in A$

so for given $y$,

Let $d(x,y)+d(y,z)=s$

We need to prove that $$d(x,q)+d(q,z)<a$$, Now we are going to work backwards, try to estimate this and find an $r$ such that the above inequality gets satisfied for any $q \in B(y,r)$

$$d(x,q)+d(q,z) \leq d(x,y)+d(y,q)+d(q,y)+d(y,z)$$ $$=d(x,y)+d(y,z)+2d(y,q)$$ $$=s+2d(y,q)$$ We want this to be less than $a$, so $2d(y,q)+s<a$

Which means $d(y,q) < \frac{a-s}{2}$

So if it happens that distance between $y$ and $q$ is less than this($\frac{a-s}{2}$) ,then q belongs to $A$. So, take your $$r=\frac{a-s}{2}$$ i.e$$r=\frac{a-d(x,y)-d(y,z)}{2}$$

Now you can reverse this process and verify this $r$ works...

But you tried proving $d(y,q)<a$, which is not you intended to prove... it is a mistake...

If you have any doubts, Let me know, I will try to help.

This kind of Working backwards helps in many places in Analysis, This is what proper analysis and estimates means... Learn this trick well...If you want to learn such kind of Analysis, let me know in the comments, I will show you...