Question

Let X be a metric space ,then show that

(a) A subset F of X is closed iff F^c is open .

(b) A subset E of X is open iff E^c is closed .

Answer 1

(a) By the definition, $F$ is a closed subset of $X$, iff any accumulation point of $F$ belongs to $F$. If $a\notin F$ then let $r=\inf\{d_X(a,x):x\in F\}$. Since $d_X(a,x)\geq0$ for all $x$, it must be $r\geq0$.

Suppose $r=0$. Then $\inf\{d_X(a,x):x\in F\}=0$ and thus, $\forall n\in\mathbb{N}$ there exists $x_n\in F$ such that $d_X(a,x_n)<1/n$. Therefore, $\lim\limits_{n\to+\infty}x_n=a$ and either $a\in F$ or $a$ is an accumulation point of $F$. Since $F$ is closed, in both cases $a\in F$, contrary to the condition that $a\notin F$.

Therefore, $r=\inf\{d_X(a,x):x\in F\}>0$. In particular, this means that for all $x\in F$ $d_X(a,x)\geq r$. Thus, the open ball $B_r(a)$ does not contain any point of $F$. Therefore, $B_r(a)\subset F^c$, and we proved that $F^c$ contains any its point $a\in F^c$ with some open ball of non-zero radius. This means that $F^c$ is an open subset of $X$.

Conversely, let it be given that $F^c$ is an open subset of $X$. We will prove that $F$ is a closed subset of $X$, that is, any accumulation point of $F$ belongs to $F$.

Let's assume the contrary: there exists a point $a$ which is an accumulation point of $F$ but $a\notin F$. Since $a\in F^c$ and $F^c$ is an open subset of $X$, an open ball $B_r(a)$ lies inside $F^c$ for some $r>0$. That is, $d_X(a,x)\geq r$ for all $x\in F$. But since $a$ is an accumulation point of $F$, there exists a sequence $x_n\in F$ ($n\in\mathbb{N}$ ) that converges to $a$. Thus, $\lim\limits_{n\to+\infty}d_X(a,x_n)=0$, which contradicts to the inequality $d_X(a,x_n)\geq r$. Therefore, the assumption is not valid. This implies that $F$ is a closed subset of $X$.

(b) The assertion "A subset E of X is open iff E^c is closed" is equivalent to "$E^c$ is closed iff $E$ is open." Since $E=(E^c)^c$, this is a consequence of the assertion of part (a).