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Reading a textbook of mine, I've encountered with a simple property and I couldn't prove it is true, I would like if someone could show me why the statement holds so I'll textually copy it.

Statement

"Let $\Omega$ be any subset of $\mathbb C$ and suppose $\alpha$ is in the interior of $\Omega$. We can, therefore, choose a positive number $\rho$ such that $B(\alpha;\rho) \subset \Omega$; it readily follows that there is a point $\xi$ in $\Omega$ with $|\xi|>|\alpha|$. To state this another way, if $\alpha$ is a point in $\Omega$ with $|\alpha|\geq |\xi|$ for each $\xi$ in the set $\Omega$, then $\alpha$ belongs to $\partial \Omega$."

All I know is that if $|\alpha-\xi|<\rho$, then $\xi \in \Omega$, how it "readily follows" that there is some $\xi$ in $\Omega$ with $|\xi|>|\alpha|$?

user156441
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2 Answers2

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Draw the picture for this. There's a ball of radius $\rho$ around $\alpha$.

Draw the line connecting $0$ to $\alpha$ and extending past that. Take a segment of length $\rho/2$ past $\alpha$, and consider the endpoint. This has larger distance from $0$, so larger absolute value - and it's still in the ball around $\alpha$, so in $\Omega$.

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If $\alpha=0$, there is nothing to show (every other point in the ball has larger magnitude).

Otherwise, draw the line through $0$ and $\alpha$. A segment of this line is a diameter of the ball centered at $\alpha$. One half of this segment contains points $\xi$ closer to $0$ ($|\xi|<|\alpha|$), and the other half contains points $\xi$ farther away from $0$ ($|\xi|>|\alpha|$).

MPW
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