[Math] Proof that every subset of a discrete space is open and closed (visualizing ‘Accumulation Points’ definition)

analysisgeneral-topologyreal-analysis

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?

Best Answer

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.

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