# Showing set $A = [0,1)$ in a discrete metric is open and closed

Let $X = \mathbb R$ with metric $\rho_0 : X \times X \to \mathbb R$ defined by $$\rho_0 (x,y) := \begin{cases} 1, & x \ne y;\\ 0, & x = y; \end{cases}$$

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$\bullet$ If $A = [0,1)$, then show that $A$ is closed and open in $\mathbb R$

- I know that all sets in discrete metric are both open and closed. They are open because if one takes a ball $B_r (x)$ for a $x \in X$ one will invariably find that the only other points contained in the ball besides $x$ are $x$ itself. And we know that singletons are usually open sets. And then the obvious thing is an entire set of singletons will mean that every complement is open. And thus the set is also closed.

But my concern is using the given interval $A$ to show it. I don't think I can visualize this specific interval. Is it the same as saying $A$ is the set $[ \rho_0 (x,x), \rho_0 (x,y))$

I just need to understand this. I hope you can help!

The next part of question is:

$\bullet$ Produce a set which is not open and closed in $\mathbb R$

And of course I cannot even begin to fathom what this may entail.

• What do you mean by "singletons are usually open sets"? – Alp Uzman Feb 25 '15 at 18:29
• @uzman I don't know if I'm stating it correctly. But I've always thought that singletons in the discrete space are open sets. Or did I misunderstand? – Siyanda Feb 25 '15 at 18:32
• And in the second one do we consider the real line with discrete metric? – Alp Uzman Feb 25 '15 at 18:34
• @Siyanda: I think your understanding is fine. To my knowledge, singletons in the discrete metric are open. – Sujaan Kunalan Feb 25 '15 at 18:36
• With discrete metric singletons are not usually open, they are open. It is not a possibility that there hides some singleton not open deep in the space :). Further, in general singletons need not be open. (The only problem in your statement was 'usually': in mathematics we don't usually like vagueness.) – Alp Uzman Feb 25 '15 at 18:40

The interval $A=[0,1)$ is defined as: $$[0,1):=\{x\in\mathbb{R}\mid 0\leq x<1\}$$ (Which is a subset of $\mathbb{R}$). Any singleton $\{x\}$ is open in the discrete metric, because for instance we can take the ball $B_{1/2}(x)$ to contain only $x$ (not just any $r$ will work). Now since unions of opens are open (axiom in topology), we find (works for any subset $A$): $$A=\cup_{x\in A}\{x\}$$ To be open. Now of course the complement of $A$ will be open by the same reasoning, hence $A$ is closed as well (I think you already understood that but formulated it a bit awkwardly).
For the second question I assume you mean $\mathbb{R}$ with the (standard) Euclidean topology/metric. Then $A$ is not open and not closed: There does not exist an open ball around $0\in A$ which is completely contained in $A$, and the same holds for $1\in\mathbb{R}\backslash A=(-\infty,0)\cup [1,\infty)$.
Although it is not wrong to write $A=[\rho(x,x),\rho(x,y)[$, this notation is not insightful I think.
To visualize $A$ with respect to the discrete metric you need to imagine that the whole set of real numbers is dispersed around. I like to think that we have a line at first (this is our "intuition" imagining real numbers as a line) and then it explodes (so that no two points are close to each other). Observe that this precisely means that singletons are open balls (and vice versa) wirh respect to discrete metric.
The reason why this visualization is legitimate is that a metric basically dictates how the space looks like, so to speak. Hence initially what we have regarding the real numbers is nothing more than an ordering (you know which number is bigger than which number). But we don't know where the numbers stand with respect to each other (e.g., $x<y\implies$ $x$ is to the left of $y$), unless we are provided with a metric.