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?
Why is the unit disk closed?
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5Follows from the defn of closed. For instance, the complement is open (why?) – 2011-08-13
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1The question http://math.stackexchange.com/questions/30039/a-subset-g-of-rn-is-open-iff-the-complement-of-g-is-closed might be helpful for reviewing properties of closed and open sets. – 2011-08-13
2 Answers
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?)
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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1What do you mean in the first sentence of the second paragraph? I'm guessing you mean that the closure of $\{y:d(x,y)<1\}$ need not be $\{y:d(x,y)\leq 1\}$ in general. You need to be more specific in your counterexample, because the metric is not determined by the topology. I guess you are referring to the *metric* $d(x,y)=1$ if $x\neq y$. What you say about $T_1$ is incorrect; every metric space is Hausdorff, hence $T_1$. – 2011-08-13
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0What I am trying to say is that in general the closure of the ball is not the closed ball. – 2011-08-13
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2In 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 $\{y:d(x,y)<1\}$. – 2011-08-13
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2@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. – 2011-08-13
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0He? Why was the answer downvoted? Sorry but in any topological space which does not have the $T_1$ the closure of a ball is not the same as the closed ball. Please look up "discrete topology" in wikipedia or read "Counterexamples in Topology". – 2011-08-13
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0@user786: What you are saying about $T_1$ is incorrect. The "balls" you are referring to are presumably for metric spaces (what is a "ball" in an arbitrary topological space?), and **every metric space has the $T_1$ property.** All it means for $\{y:d(x,y)\leq 1\}$ to not be the closure of $\{y:d(x,y)< 1\}$ is that there exist $z$ such that $d(x,z)=1$ but $z$ is not an accumulation point of $\{y:d(x,y)< 1\}$. This can happen in metric spaces much nicer than discrete spaces. – 2011-08-13
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2Well, 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](http://rationalwiki.org/wiki/Not_even_wrong). – 2011-08-13
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0By the way, I did not downvote because it may help the OP to think of how the closed unit disk in the plane is the closure of the open unit disk. I do not upvote because there are several things that are misleading in this answer, as I have tried to indicate in my previous comments. – 2011-08-13
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1Ohhh my bad, I am confusing the discrete topology with something else :(. Sorry about the misunderstanding. Might be too late to do math... – 2011-08-13
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2Okay, 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](http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/27265#27265). 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. – 2011-08-13
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5Don't worry about the comments. At least I learned something. – 2011-08-13