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Prove that any set consisting of finitely many points are bounded. (complex plane)

I am not sure how to prove this?

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The set is contained in any disk centered at the origin of radius bigger than the maximum of the absolute values of the points. Think geometrically. – anon Jan 21 '13 at 3:53
@anon But why would it be bounded? Shouldn't it be an open set? Can you explain to me why it is bounded instead of being an open set please? – Q.matin Jan 21 '13 at 3:54
@Q.matin, what is your definition of "bounded set in the complex plane"? Anon's comment can hardly be clearer. – DonAntonio Jan 21 '13 at 3:56
@Q.matin "boundary point" has nothing to do with "bounded". – Ted Jan 21 '13 at 3:59
A subset of the plane (or a metric space in general) is said to be bounded, or of finite diameter, if it is contained in some disk of specified (so, finite) radius. – anon Jan 21 '13 at 4:02
up vote 1 down vote accepted

If $\{z_1,z_2,\cdots,z_n\}$ is the finite collection of complex numbers, then $max(|z_1|,|z_2|,\cdots,|z_n|)$ lies in the set $\{|z_1|,|z_2|,\cdots,|z_n|\}$. Therefore, for any $k>max(|z_1|,|z_2|,\cdots,|z_n|)$, $|z_i|\lt k$ $\forall i\in \{1,2,3,\cdots,n\}$. Thus the set is bounded.

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Thanks a lot, Avatar!! – Q.matin Jan 21 '13 at 4:09

Let $A=\{a_1,a_2,\cdots,a_n\}$, $r=\max\limits_{1\le i\le n}|a_i|$. Then $A\subset B(0,2r)$. ($B(z,R)$ is open ball whose center is at $z$ and radius $R$.)

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Thanks, a lot!! – Q.matin Jan 21 '13 at 4:05

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