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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$\begingroup$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?)
$\endgroup$ $\begingroup$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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