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I'm working through Rudin's "Principles of Mathematical Analysis" on my own, so I don't want the full answer. I'm only looking for a hint on this problem.

Rudin states without proof that the set $X = \{z \ \text{complex}: |z| \lt 1\}$ is not closed. I'm having trouble rigorously showing that this is true, even though I'm fairly confident that because $1 + 0i$ is a limit point of this set and is obviously not in the set, the set isn't closed.

To figure this out, I tried to show that the similar set $E = \{z \ \text{complex}: |z| \leq 1\}$ is closed. My hope is that by proving this, I'll be able to see how to show that the previous one isn't closed. Here's my thought process so far, using the definitions discussed previously in Rudin.

  1. $p$ is a limit point of E, so for every $N_r(p)$, $\exists q \in N_r(p)$ such that $q \ne p$ and $q \in E$.
  2. $q \in N_r(p) \Rightarrow |p - q| < r \Rightarrow |p-q| = r-h$, where $ 0 < h < r$ by definition of nbhd.
  3. $q \in E \Rightarrow |q| \leq 1$.
  4. I can use the triangle inequality to show that

\begin{align} |p-q| &= r - h \\ &\leq |p| + |q| \\ &\leq |p| + 1 \\ \end{align}

but I'm stuck on how to proceed. Basically, to show that E is closed, I need to show that if $p$ is a limit point of E, then $p \in E$, i.e. $|p| \leq 1$. I'm hoping that continually solving these examples and exercises will give me a better understanding of how to approach these problems, because now I just feel like my strategy is "write down everything I know that seems related to the problem and see what fits together."

Any hints?

M T
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    First, +1 for a question showing excellent work and preparation. Here's a hint: you may find it useful to try to show that the distance function is continuous and then use that insight to prove the openness/closedness of the open/closed unit balls. – Neal Nov 28 '14 at 05:16
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    @Neal Thanks for the hint; I'll see what I can do with this. I must be missing some background, though, because Rudin's book hasn't discussed continuity yet (and won't for another two chapters), so I'm not sure how to prove something is continuous. – M T Nov 28 '14 at 14:34

3 Answers3

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Use proof by contradiction.

Define $Q=\{z\in{\mathbb C}:|z|\leq1\}$

If $p$ is a limit point of $Q$ and $|p|>1$ then $...$

let $t = |p| - 1$. There is an open ball $S_t(p)$ of radius $t$ around $p$ which does not contain any points of $Q$ (that is, $Q\cap S_t(p)=\emptyset$). This contradicts the assumption that $p$ is a limit point of Q. Thus there cannot be any limit points of $Q$ with $|p|>1$.

(Sorry if that is too much hint for you.)

Suzu Hirose
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  • @learnmore What are you asking for, the definition of the open ball, or why it doesn't intersect $Q$? – Suzu Hirose Nov 28 '14 at 06:55
  • @learnmore I have no idea what kind of answer you want to hear. – Suzu Hirose Nov 28 '14 at 08:02
  • @SuzuHirose Here's my thought process. Thinking about the nbhd $S_t(p)$ (well, the open ball, but that's just a special case of a nbhd), I know that it consists of all points $y$ such that $|y-p|<t$ and that I need to show that $|y| > 1, \forall y \in S_t(p)$, but I'm still struggling with that. I'll think about it some more and come back. – M T Nov 28 '14 at 19:28
  • @Michael I don't know what definition of "limit point" you are using but the fact that there is a neighbourhood of $p$ which doesn't contain any elements of $Q$ should be enough, I don't think you need to prove anything about $y>1$. – Suzu Hirose Nov 28 '14 at 22:37
  • Because assume there is a point $r\in S_t(p)$, therefore $|r|=|r-p+p|>|r-p|+|p|>t+1>1 \Rightarrow r \notin Q$ – K.K.McDonald Apr 27 '20 at 15:19
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I'm not too big on the "hints" approach myself, and I'm not sure I'm too good at it, but I do think this a good question, so I'll try not to say too much. And I'll ask our OP Michael to let me know how it works out.

Think of this: the map $\vert \cdot \vert: \Bbb C \to \Bbb R$ is continuous. To see this, recall that the triangle inequality implies that

$\vert \vert w \vert - \vert z \vert \vert \le \vert w - z \vert; \tag{1}$

thus if $\vert w - z \vert < \epsilon$, $\vert \vert w \vert - \vert z \vert \vert < \epsilon$; that's continuity. Now, continuity means that the inverse image of an open set is . . . ? And the inverse image of a closed set is . . . ?

OK, I'll stop typing here for the moment. If more is needed, let me know.

Hope this helps (but not too much). Cheers,

and as ever,

Fiat Lux!!!

Robert Lewis
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  • I guess I'm still confused because these examples are given, without proof (and usually Rudin sketches a proof, at least in the earlier chapters) before the middle of Chapter 2. The definition of continuity and proofs using it don't occur until Chapter 4, but should I be able, in the middle of Chapter 2, to work out a proof that apparently uses material from much further on in the book? I guess my maths understanding isn't good enough yet, unfortunately. – M T Nov 28 '14 at 18:57
  • @Michael: well, as I said, I'm not too good at the hint game, so if I overdid it in terms of sophistication, please forgive me; that wasn't really my intention. I'm not sure how Rudin's book defines open and closed sets. but if he uses the definition that $Y$ is closed iff the complement of $Y$, $Y^c$, is open, then using equation (1) from my answer, you can see that if $\vert z \vert > 1$ and $\vert w - z \vert < \epsilon$ for sufficiently small $\epsilon$ (so that $w$ is in the open $\epsilon$-ball or -disk around $z$), then $\vert w \vert$ > 1 as well. To be continued . . . – Robert Lewis Nov 28 '14 at 19:47
  • Hmmm, it might be that he provides these examples before the major discussion of open and closed sets, because I think those discussions occur a few pages after. I might have a better understanding of the necessary tools if I finish the chapter and then return to those examples. – M T Nov 28 '14 at 19:53
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    @Michael: (continuation of previous comment) This means that the set of $z$ with $\vert z \vert > 1$ is open, since every point in in contains an open disk also in it; thus its complement, that is, ${ z \mid \vert z \vert \le 1 }$ is closed. This argument is similar in spirit to Suzu Hirose's; when you learn more about continuity, you'll see how the argument hinted at in my answer works, and that it is essentially the same as the one I've put into these comments. I do hope these words clarify somewhat; if not, let me know . . . Cheers! – Robert Lewis Nov 28 '14 at 19:54
  • @Michael: I think I might be done with the hint game for awhile . . . ;) But I'm sure you'll get all this straightened out before too long. Best of luck with it all! – Robert Lewis Nov 28 '14 at 19:57
  • Thank you very much for the help; that does make some of the concepts a lot clearer, so I'll come back and post an update on how I solved it once I do. Thank you again! – M T Nov 28 '14 at 20:40
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Assume that ${\bar{D}}(0,1)$ be the set of complex numbers $z$ such that$|z|\leq1$, and let ${\bar{D}}(0,1)^{c}$ be its complement in the complex plane. To prove that ${\bar{D}}(0,1)$ is closed, one can prove that ${\bar{D}}(0,1)^{c}$ is open. (The two statements are equivalent). So, we shall prove that ${\bar{D}}(0,1)^{c}$ is open. To show that this is so, it suffices to prove that every point $w\in{{\bar{D}}(0,1)^{c}}$ must be an interior point of this set. To prove this, since $|w|>1$, let ${\epsilon}>0$ be chosen so that $0<{\epsilon}<(|w|-1)$. We contend that the open disc $D(w,{\epsilon})$ of radius $\epsilon$ centered at $w$ must be a subset of ${\bar{D}}(0,1)^{c}$. If we prove this statement, since the argument applies to an arbitrary $w\in {{\bar{D}}(0,1)^{c}}$, we could then conclude that every point of ${\bar{D}}(0,1)^{c}$ must be an interior point of ${\bar{D}}(0,1)^{c}$, proving that this set must be an open set, and hence its complement ${\bar{D}}(0,1)$ must be a closed set. So, assume that $z\in{D(w,{\epsilon})}$. Then, $$|z|=|w+(z-w)|\geq|w|-|z-w|>|w|-(|w|-1)=1.$$ Thus, if $z\in{D(w,{\epsilon})}$, then $|z|>1$, and so $z\in{{\bar{D}}(0,1)^{c}}$. Hence, ${D(w,{\epsilon})}\subset{{{\bar{D}}(0,1)^{c}}}$. Since this argument applies to every point $w\in{{{\bar{D}}(0,1)^{c}}}$, we conclude that every point of this set must be an interior point, and hence ${{{\bar{D}}(0,1)^{c}}}$ must be an open set, proving that ${\bar{D}}(0,1)$ must be a closed set. This proof only uses material from Chapter 2 of Rudin's textbook. If you want a simpler argument, observe that the mapping $z\mapsto{|z|}$ is a continuous mapping ${\bf{C}}\rightarrow{}[0,\infty]$, and ${\bar{D}}(0,1)$ being the inverse image of the closed subset $[0,1]$ of $[0,\infty]$ under this mapping must be a closed subset of the complex plane.

student
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