So I am going through "Understading Analysis" by Abbott and have simple question about one of his examples.
He states the following:
"To see that$ \ (c,d)\ $is open in the sense just defined [above], let $ \ x\in(c,d) \ $be arbitrary. If we take $ \ \epsilon=min\{x-c,d-x\}, \ $then it follows that $V_{\epsilon}(x)\subseteq (c,d).$ It is important to see where the argument breaks down if the interval includes either one of its endpoints."
My question:
If you are taking $ \ \epsilon=min\{x-c,d-x\}, \ $ I would think that your $V_{\epsilon}(x)$ actually would have some outside either $c$ or $d$ depending of what is the minimum since you are taking $x$ minus the one of the end points.
Just to clarify my confusion: If you take the open interval $(1,3)$ and take an arbitrary $x \in (1,3)$ then I would think that with $ \ \epsilon=min\{x-1,3-x\}, \ $ the $V_{\epsilon}(x)$ would in fact have points outside of $(1,3)$ since we have created an epsilon using the endpoints.
I am not sure if I am being clear of what I am missing, but would appreciated if someone could clarify!
If $\epsilon\le x-1$ then $x-\epsilon\ge x-(x-1)=1$. Also, if $\epsilon\le 3-x$, then $x+\epsilon\le x+(3-x)=3$. Thus, for $0<\epsilon\le\min\{x-1,3-x\}$, we have:
$$1\le x-\epsilon\lt x+\epsilon\le 3$$
so the interval $V_\epsilon(x)=(x-\epsilon, x+\epsilon)$ is contained in $(1,3)$. You can generalise this to any $c, d$ instead of $1,3$.