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Manufacture a topology on $\mathbb{N}$ such that the boundary of finite intervals consists of its extreme points.

The answer to this question correctly says this is impossible. But I've set myself the challenge of doing it anyway. Can this this impossibility be surmounted by assigning each integer a pair of points, one arbitrarily less than the other?

This is partly motivated by dissatisfaction with the conventional discrete topology of $\mathbb{N}$, partly as an exercise in learning topology concepts and testing my understanding of the subject, but I've also got it into my head that something similar to this might be of utility in the Parity Problem. But that's just background... on to the actual question.

Let any integer $n$ be the union of two points $n_l,n_r$ such that $n=\{n_l\}\cup \{n_r\}: n_l\leq n_r$ and $n_r-n_l=\epsilon$

Now if the desired boundary of $A=\{a,a+1,a+2,...,a+n\}$ is to be $\{a,a+n\}$ (which is impossible under conventional rules) we actually have in this new sense its boundary is $\{\{a_l\},\{a+n_r\}\}$ and its interior is $\{\{a_r\},a+1,a+2,...,\{a+n_l\}\}$:

Does this achieve the stated aim of a topology on $\mathbb{N}$ in which the endpoints of a sequence of integers are the boundary? Is this anything more than pointless exercise?

it's a hire car baby
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1 Answers1

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As you describe it, it is not a topology on $\Bbb N$ because the base set is not $\Bbb N$. However, your set is actually $\Bbb N$ in disguise, as it consists of countably many discrete points in a total order. We can make a bijection by taking $n_l \leftrightarrow 2n, n_r \leftrightarrow 2n+1$. The difference in spacing between $\epsilon$ and $1-\epsilon$ does not matter because we have not talked about a metric. You are then trying to define a basis of open sets as $[2k+1,2m]$ for $m \gt k$. The closure of the topology under unions requires that $(-\infty,2k]$ and $[2m+1,\infty)$ must be open as well, so $[2k+1,2m]$ is closed and therefore clopen. We therefore get the discrete topology with the points being $[2k+1,2k+2]$. It is not $T_0$ because there is no way to distinguish $2k+1$ from $2k+2$, or in your original discussion, $n_r$ from $(n+1)_l$

Ross Millikan
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  • Thanks. The question asks to make a topology on $\mathbb{N}$ such that the boundary of finite intervals consists of its extreme points so perhaps the answer is to augment the above with a metric whereby $d: \mathbb{N}l\times\mathbb{N}_r\to\mathbb{N}{>0}$ or pseudometric in which $d(x_l,x_r)=0$. Perhaps this gets closer to the objective? – it's a hire car baby Aug 05 '17 at 18:22