# Thread: Prove a subset of discrete metric is both open and closed

1. ## Prove a subset of discrete metric is both open and closed

Problem: Let X be any set considered a metric space with the discrete metric. With this metric, show that every subset of X is both open and closed in X.

2. This sort of proof is hard to explain without knowing exactly what your particular definitions are. However, here is some general guidance.
Because this is the discrete metric $\left( {\forall t \in X} \right)\left[ {B_{1/2} \left( t \right) = \{ t\} } \right]$. From that, it is easy to see that there are no limit points in this space. By default every set contains its limits points and is therefore closed. Likewise, the complement of any set is closed so the set is open.

3. What does it mean a discrete metric?
$d(x,y) = 0$ iff $x=y$ otherwise $d(x,y)=1$ iff $x\not = y$.?

4. Originally Posted by ThePerfectHacker
What does it mean a discrete metric?
$d(x,y) = 0$ iff $x=y$ otherwise $d(x,y)=1$ iff $x\not = y$.?
That is the standard definition.

5. Here is what I have so far, with the current definitions that I have.

Proof: Let X be any set, the discrete metric $d(x,y)= \left\{\begin{array}{cc}1,&\mbox{ if }
x \not = y\\0, & \mbox{ if } x=y\end{array}\right.$

We want to show that for any subset, say $O \subset X$, is open. (i.e. $N_{r}(O) = O$ for some $r > 0$)

Pick $r = \frac{1}{2}$, for a point $y \in O$, $N_{\frac{1}{2}}(y) = \{x \in X : d(x,y) < \frac{1}{2} \}$

Let $z \in N_{\frac{1}{2}}(y)$ , $d(z,y) < \frac{1}{2}$, and since d is discrete,

$z=y \Rightarrow z \in O \Rightarrow N_{\frac{1}{2}}(y) \subset O \Rightarrow N_{\frac{1}{2}}(O) \subset O \Rightarrow O = int(O)$,

thus proves $O$ is open.

Now we will show that $O$ is closed.

Pick $r=2$, then for a point $y \in O$, $N_{2}(y) = \{x \in X : d(x,y) < 2 \}$

Let $z \in N_{2}(y)$, then $d(z,y) < 2$

Now, since d is discrete, $z \not = y \Rightarrow z \in X \Rightarrow N_{2}(O)=X$

Thus $O$ is closed.

Q.E.D.

Is this convincing?

Now we will show that $O$ is closed.
Pick $r=2$, then for a point $y \in O$, $N_{2}(y) = \{x \in O : d(x,y) < 2 \}$
Let $z \in N_{2}(y)$, then $d(z,y) < 2$
Now, since d is discrete, $z \not = y \Rightarrow z \in X \Rightarrow N_{2}(O)$
Thus $O$ is closed.
Is this convincing?
It is not convincing at all. It is wrong.
There are basically two ways to show that a set is closed in a metric space.
If the set contains all its limit points it is closed.
If the complement of the set is open then the set is closed.
Frankly, I have no idea what you are aiming at in the above.

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