Show that the set of all complex numbers $z$ such that $|z| \leq 1$ is closed? 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. 


*

*$p$ is a limit point of E, so for every $N_r(p)$, $\exists q \in N_r(p)$ such that $q \ne p$ and $q \in E$.

*$q \in N_r(p) \Rightarrow |p - q| < r \Rightarrow |p-q| = r-h$, where $ 0 < h < r$ by definition of nbhd.

*$q \in E \Rightarrow |q| \leq 1$.

*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?
 A: 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.)
A: 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!!!
A: 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.
