# Proof that every subset of a discrete space is open and closed (visualizing 'Accumulation Points' definition)

I know this question is a duplicate, but I want to understand it in terms of accumulation points and internal points, etc.

QUESTION

Let X be any set and $d : X \times X \to \mathbf{R}$ be given by $$d(x,y) = \begin{cases} 0, & \text{if x = y} \\ 1, & \text{if x \neq y} \\ \end{cases}$$

Prove that every subset of $X$ is both open and closed.

REMARKS

First of all the definition I like to use of 'closed' is the following:

Let $E$ be a set of real numbers. The set $E$ is said to be closed provided that every accumulation point of $E$ belongs to the set $E$

Accumulation point: Any point $x \in R$ (not necessarily in $E$) provided that for every $c>0$ the intersection $(x-c,x+c)\cap E$ contains infinitely many points.

Thomson, Bruckner, Bruckner Elementary Real Analysis, 2nd Edition (2008)

I began to visualise the subsets and exactly what they would look like.

I considered case 1 - where $d(x,y) = 0 for x=y$

I imagined this would be a group of subsets that looked like this:

$(..., [x_{i-2}], [x_{i-1}], [x_{i}], [x_{i+1}],...)$

Basically, sets of single values. Now, I begin to think of accumulation points. And how I could work that into this particular situation.

I thought of a literal example. Let's look at a interval $[x_i - c, x_i + c]$ This point would only intersect with our $E$ (in this case the set $[x_i]$) when $c=o$. So then how can we say this set is one that contains infinitely points and satisfies our definition?

Am I right in thinking that this set only has one accumulation point? And is that all I need to do to say that this set is closed?

Or am I just looking at all of this the wrong way?

• The approach utilising the definition of openness in metric spaces may be a little bit more straightforward. Since $$B_{\frac{1}{2}}(x) = \{x\} \qquad \forall x\in X$$ All one-point sets are trivially open. All subsets are unions of those and so on... For the closedness you can WLOG (why?) assume you sequence is the constant sequence and there you go... Commented Aug 26, 2013 at 19:29
• What is WLOG? @AlexR Commented Aug 26, 2013 at 19:41
• "without loss of generality" Commented Aug 26, 2013 at 19:42

Here are the definitions that will give you the answers relatively quickly:

Definition (opennes)
Let $(X,d)$ be a metric space and $A\subset X$. Then $A$ is open iff $$\forall a \in A\quad \exists \delta > 0: \ B_\delta (a) := \{x\in X: d(x,a) < \delta\} \subset A$$

and

Definition (closedness)
Let $(X,d)$ be a metric space and $(a_n)_{n=1}^\infty \subset A$ with $$\lim_{n\to\infty} d(a_n,a) = 0$$ If this implies $a\in A$, then $A$ is closed.

The problem is that you were taking $[x_i - c, x_i + c]$ from the book to be your open ball of radius $c$ about $x_i$, and this is with respect to the standard metric on $\mathbb{R}$, which is $d(x,y) = |x-y|$. However, your metric is different, so if you take a ball of any radius about $x$ with respect to your weird metric, you will see what AlexR is talking about.

In any case, the point is: The book reference has a particular metric in mind when defining accumulation point. You need a general one that works for all metric spaces.