The complement of a finite set in $\mathbb{R}$ is open?

Question

I was reading " Notes on Introductory Point - Set Topology" by Allen Hatcher, Chapter 1 as he gave as an example of an open set : " the complement of a finite set in $\mathbb{R}$ ". I want to prove this but I do not know how like if I have the following closed set $\{1, 2 ,3 \},$ how to prove that its complement is open?

Answer 1

Let $X=\mathbb R\setminus\{1,2,3\}$. Let $x\in X$ be given. Take $\delta=\min\{|x-1|,|x-2|,|x-3|\}$ i.e. the minimum distance from $x$ to one of the excluded points. Then $\left(x-\frac\delta2,x+\frac\delta2\right)$ contains $x$ and is completely contained inside $X$. Since $x\in X$ was aribitrary, $X$ is open.

There's nothing magic about the set $\{1,2,3\}$ in this proof -- the same argument could be applied to any finite set.

Answer 2

Well, every finite set is close (with the Euclidean metric) because a finite set has no limit points.

Pf: If $A$ is finite then for any $x\in \mathbb R$ there are only finitely many $d(x,a)$ where $a\in A$ and $a \ne x$ so a number $r = \min d(x,a)> 0; a \in A; a\ne x$ exists and $B_r(x)$ has no points of $A$ other than $x$, so $x$ is not a limit point of $A$.

And a set is closed if and only if its complement is open.

Pf: Standard exercise. If $E$ is closed and $a\in E^c$ then $a$ is not a limit point of $E$ so there is a neighborhood of $a$ with no points of $E$ other than $a$, but $a$ isn't in $E$ so that neighborhood is entirely id $E^c$ so $E^c$ is open. Likewise if $E^c$ is open then for every point, $a$ in $E^c$ there is a neighorhood entirely in $E^c$ so $a$ can not be a limit point of $E$ and therefore limit points of $E$ can not be outside of $E$.

.... that's it....

But if you want to prove it is open directly, well, Matthew Daly's answer does that.

If $x \not \in A$ then as there are finite $a \in A$ there are finite $d(x,a)$ and as $x \not \in A$ no $a =x$ and all $d(x,a) > 0$ and there is a minimum $d(x,a)$. Call it $r$. The $B_r(x) \subset A^c$ so $A^c$ is open.

Answer 3

$\Bbb R$ is a metric space. Metric spaces are always $T_1$. Thus points are closed. But the finite union of closed sets is closed. Thus the complement of a finite set is open.

Answer 4

In a metric space the complement of any finite set is open.

Proof:

Let (X,d) be a metric space.

Let F be a finite subset of X.

Let F={x1,...,xn}.

Claim:

X-{x1,...,xn} is an open set.

Let y∈X-{x1,...,xn}.

⇒ y∈X and y∉{x1,...,xn}.

⇒ y∈X and ∀i∈{1,...,n}(y≠xi).

⇒ ∀i∈{1,...,n}(d(y,xi)≠0).

Let ∀i∈{1,...,n}(d(y,xi)=ri).

Let r=min{ri:i∈{1,...,n}}.

⇒ ∀i∈{1,...,n}(d(y,xi)<r).

Claim: B(y;r)⊆X-{x1,...,xn}.

Clearly ∀i∈{1,...,n}(xi∉B(y;r)).

Let if possible,

∃i∈{1,...,n}(xi∈B(y;r)).

⇒ d(y,xi)<r and d(y,xi)=ri.

⇒ ri<r.

Contradiction to the fact that r=min{ri:i∈{1,...,n}}.

Therefore,

∀i∈{1,...,n}(xi∉B(y;r)).

⇒ B(y;r)∩{x1,..., xn}=Φ.

⇒ B(y;r)⊆X-{x1,...,xn}.

⇒ ∀y∈X-{x1,...,xn}(∃r>o(B(y;r)⊆X-{x1,...,xn})).

⇒ X-{x1,...,xn} is an open set.