I am reading "Analysis on Manifolds" by James R. Munkres.
In Step 2, the author uses the two balls $B(b,2\delta)$ and $B(b,\delta)$.
I think only $B(b,2\delta)$ is sufficient for Step 2.
Why does the author use the two balls?
I wonder why the author didn't write as follows:
My Step2. We show that the set $B$ is open in $\mathbb{R}^n.$ Given $b\in B,$ we show $B$ contains some open ball $B(b;\delta)$ about $b.$
We begin by choosing a rectangle $Q$ lying in $A$ whose interior contains the point $a=f^{-1}(b)$ of $A.$ The set $\operatorname{Bd} Q$ is compact, being closed and bounded in $\mathbb{R}^n.$ Then the set $f(\operatorname{Bd}Q)$ is also compact, and thus is closed and bounded in $\mathbb{R}^n.$ Because $f$ is one-to-one, $f(\operatorname{Bd}Q)$ is disjoint from $b;$ because $f(\operatorname{Bd}Q)$ is closed, we can choose $\delta>0$ so that the ball $B(b;\delta)$ is disjoint from $f(\operatorname{Bd}Q).$ Given $c\in B(b;\delta)$ we show that $c=f(x)$ for some $x\in Q;$ it then follows that the set $f(A)=B$ contains each point of $B(b;\delta),$ as desired. See Figure 8.1.
Given $c\in B(b;\delta),$ consider the real-valued function $$\phi(x)=||f(x)-c||^2,$$ which is of class $C^r.$ Because $Q$ is compact, this function has a minimum value on $Q$; suppose that this minimum value occurs at the point $x$ of $Q.$ We show that $f(x)=c.$
Now the value of $\phi$ at the point $a$ is $$\phi(a)=||f(a)-c||^2=||b-c||^2<\delta^2.$$ Hence the minimum value of $\phi$ on $Q$ must be less than $\delta^2.$ It follows that this minimum value cannot occur on $\operatorname{Bd}Q,$ for if $x\in\operatorname{Bd}Q,$ the point $f(x)$ lies outside the ball $B(b;\delta),$ so that $||f(x)-c||\geq\delta.$ Thus the minimum value of $\phi$ occurs at a point $x$ of $\operatorname{Int}Q.$
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