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Open + Closed

Example 1. Let
(X,d)(\mathcal{X},d)
be a metric space and
xXx\in\mathcal{X}
. Prove
{x}X\{x\}\subset \mathcal{X}
is closed.
Hint
Visual
Formal Write-Up
Follow-Up
It may be easier to prove the complement is open than the prove
{x}\{x\}
is closed.
Sometimes it is hard to make a direct proof. Here we take an alternative route by proving that the complement is open. We pick a point not equal to and must show . This is done by picking a radius to be half the distance between and so that . We get the proof by rearranging the triangle inequality to show
We use the fact a set is open if and only if its complement is closed. We proceed by showing the complement set
CX{x}\mathcal{C} \triangleq \mathcal{X} - \{x\}
is open. Recall
C\mathcal{C}
is open if and only if each element of
C\mathcal{C}
is an interior point. To this end, let
zCz \in \mathcal{C}
be given. It suffices to show
zint(C)z \in \text{int}(\mathcal{C})
.
Since
zCz\in\mathcal{C}
,
zxz \neq x
, and so
d(x,z)>0d(x,z) > 0
. Thus, setting
rd(x,z)/2r \triangleq d(x,z) / 2
gives
r>0r > 0
. For all
yB(z,r)y \in B(z,r)
, the triangle inequality and choice of
rr
imply
d(x,z)d(x,y)+d(y,z)      d(x,y)d(x,z)d(y,z)2rr=r>0      yx      yC.\begin{array}{cl} &d(x,z) \leq d(x,y) + d(y,z) \\ \implies \ \ & d(x,y) \geq d(x,z) - d(y,z) \geq 2r - r = r > 0 \\ \implies \ \ & y \neq x \\ \implies \ \ & y \in \mathcal{C}.\end{array}
Thus,
B(z,r)CB(z,r) \subseteq \mathcal{C}
, i.e.
zz
is an interior point.
\blacksquare
As noted previously, when proving something holds "for all elements,'' it is typically easiest to let an arbitrary element be given and prove the relation for that element. The result then will follow.
Example 2. Let
X\mathcal{X}
be a metric space. If
X=AB\mathcal{X} = \mathcal{A}\cup \mathcal{B}
with nonempty
A\mathcal{A}
and
B\mathcal{B}
,
AB=\overline{\mathcal{A}} \cap \mathcal{B} = \emptyset
, and
AB=\mathcal{A} \cap \overline{\mathcal{B}} = \emptyset
. Show both
A\mathcal{A}
and
B\mathcal{B}
are open.
Visual
Formal Write-Up
Illustration of and a ball in . Note the ball is shown in red and is not the entire green circle. This follows from the fact a ball in can only contain elements in . Thus, in general, a ball might not be round.
Let
xAx\in \mathcal{A}
be given. Set
hoxho_x
to be the distance from
xx
to
B\overline{\mathcal{B}}
, i.e.
hoxd(x,B)=inf{d(x,y):yB}.ho_x \triangleq d(x, \overline{\mathcal{B}}) = \inf \{ d(x,y) : y \in \overline{\mathcal{B}}\}.
Since
{x}\{x\}
is compact,
B\overline{\mathcal{B}}
is closed and
{x}B=\{x\}\cap\overline{\mathcal{B}}=\emptyset
, the distance
hoxho_x
is positive.* Then consider the ball
B(x,ρx2 ){z:d(x,z)<ρx2, zX}.B\left(x, \frac{\rho_x}{2}\ \right) \triangleq \left\lbrace z : d(x,z) < \frac{\rho_x}{2}, \ z \in \mathcal{X}\right\rbrace.
For all
yBy \in \mathcal{B}
,
d(x,y)ρx>ρx2          yB(x,ρx2).d(x,y) \geq \rho_x > \dfrac{\rho_x}{2} \ \ \ \implies \ \ \ y \notin B\left(x, \frac{\rho_x}{2}\right).
Hence
B(x,ρx/2)B=B(x,\rho_x/2)\cap \overline{B} = \emptyset
. With the fact
B(x,ρx/2)X=ABB(x,\rho_x/2) \subseteq \mathcal{X} = \mathcal{A}\cup \mathcal{B}
, we deduce
B(x,ρx)AB(x,\rho_x) \subseteq \mathcal{A}
, i.e.
xx
is an interior point of
A\mathcal{A}
. Since
xx
was arbitrarily chosen, every element of
A\mathcal{A}
is an interior point, i.e.
A\mathcal{A}
is open. An analogous argument applies to show
B\mathcal{B}
is open, completing the proof.
\blacksquare
*This follows from a prior example.