# Why is the unit disk closed?

I read a question stating that if $z$ is complex, then $|z|\leq 1$ is a closed set. I think this is just saying that the unit disk is a closed set. Why is that so?

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Follows from the defn of closed. For instance, the complement is open (why?) – user17762 Aug 13 '11 at 5:22
The question math.stackexchange.com/questions/30039/… might be helpful for reviewing properties of closed and open sets. – Jonas Meyer Aug 13 '11 at 5:31

Look at the complement of the set $S = \{ z \in \mathbb{C}: |z| \leq 1 \}$. The complement is given by $S^c = \{ z \in \mathbb{C}: |z| > 1 \}$. Consider a point $z_0 \in S^c$. $|z_0| > 1$ and hence let $|z_0| = 1 + r$ where $r > 0$. Consider the ball of radius $r$ centered at $z_0$ i.e. $B_r(z_0) = \{v \in \mathbb{C}:|v-z_0| < r \}$. Clearly, $B_r(z_0) \subset S^c$. This follows from triangle inequality since $$|z_0| = |z_0 - v + v| \leq |z_0 - v| + |v| \implies |v| \geq |z_0| - |z_0 - v| > (1+r) - r = 1$$ Hence, $|v|>1 \implies v \in S^c \implies B_r(z_0) \subset S^c$. Hence, $S^c$ is open since given any point $z_0 \in S^c$ we can find a open neighborhood lying completely inside $S^c$ and hence $S$ is closed.

Equivalently, you can try to prove that the set $S = \{ z \in \mathbb{C}: |z| \leq 1 \}$ contains all its accumulation points. The proof of this is again not hard. Look at a subsequence converging to an accumulation point and prove that if you have $|z_n| \leq 1$, then $\displaystyle \left| \lim_{n \rightarrow \infty} z_n \right| \leq 1$. (Hint: If not, what will happen?)

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The unit disc $\bar D$ is the closure of the open unit disc $D$ defined as all complex numbers $z$ such that $|z|<1$. It just happens that with the standard topology of the plane, the closure of the open unit disc is the closed unit disc. So the set of all $z$ such that $|z| \leq 1$ is closed.

Note that in general the closure of the set is not the closed set. Indeed in the discrete topology this fails. What we need to insure that the closure of a ball coincide with the closed ball is the separation axiom called $T_1$. In topological spaces which do not have the $T_1$ axiom (discrete topology), this principle fails. Alternatively you can see it with the perspective Sivaram offers: check that the complement is open.

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In any case, perhaps it should be emphasized $\{y:d(x,y)\leq 1\}$ is closed in any metric space whether or not it is the closure of $\{y:d(x,y)<1\}$. – Jonas Meyer Aug 13 '11 at 5:48
@user786: Pardon me, but what does a closed ball mean to you? The closure of any set is certainly a closed set. Would that not be a closed ball to you? I agree with Jonas' remarks. – Jyrki Lahtonen Aug 13 '11 at 5:56
Well, since you ask: I was the second downvoter here (I normally wouldn't do that). But: The first sentence of the second pargraph is imprecise at least. The two sentences including $T_1$ are not even wrong. – t.b. Aug 13 '11 at 6:20
Okay, then let's save your post: You wanted to explain that the closure of $\{y\,:\,d(x,y) \lt 1\}$ is not in general $\{y\,:\,d(x,y) \leq 1\}$, it may be strictly contained (note that the latter set is closed since the metric is continuous). That's a very good point: even Terry Tao found that it is worth making. Then strip out every mention of $T_1$, just to be on the safe side and I'll be happy to remove my comments. I'm pretty sure that Jonas will do so, too. – t.b. Aug 13 '11 at 6:39
Don't worry about the comments. At least I learned something. – user786 Aug 13 '11 at 6:49