How to show a set in $ \Bbb C $ is open?

Question

Knowing that $ \vert d(z,A)-d(w,A)|\le\vert z-w|$, for all z,w$\in\Bbb C$ prove that $ A_\mathcal E = ${z:d(z,A)<$ \mathcal E$} is open.

Here is my attempt: As we want to show $ A_\mathcal E$ is open, then we should take a point on it and show that is an interior point of $ A_\mathcal E$. So let a$\in A_\mathcal E $, then d(a,A)<$ \mathcal E$, i.e. inf{x$\in A$:d(a,x)}<$ \mathcal E$.

Also we have$ \vert d(z,A)|\le\vert z-w|$ if w$\in A$.

$ \Rightarrow $ $ \ d(z,A)\le\vert z-w|$

$ \Rightarrow $inf{x$\in A$:d(z,x)} $\le\vert z-w|$

and then I don't know how to continue..

:(

Answer 1

So, good use of the triangle inequality. To show a set is open, we can just construct an open ball around every point, so just take an open ball with the smallest radius which contains all the points.

You are on the right track, but you need some clean-up: just how take the radius of your open ball to be $\epsilon=|z-w|+\delta$. Note by triangle inequality, you have $\leq$ which defines a closed ball, so you need the smallest open ball which contains points $|z-w|$. Since this open ball contains all the points of interest, you can conclude that the set $A_{\epsilon}$ is open.

Also, note that by your definition of $A_{\epsilon}$ all the points with $d(w,A)$ are contained in the ball.