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Problem

Consider the set $X= \{1,2\}×\mathbb{Z}^+$ with the dictionary order as an ordered set with a smallest element.

Doubt

  1. I am not able to understand what kind of intervals form the basis of this topology.

  2. Also I am not able to understand how this topology is not a discrete topology

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  • $\begingroup$ It is defined the way you have defined it. It says that (2,1) is an exception. $\endgroup$ – blue boy Nov 4 '18 at 7:10
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Aha, the OP means $\mathbb{Z}^+ = \{1,2,3,4,5,\ldots\}$ as the second component...

Consider what the order looks like: $ (1,1) < (1,2) < (1,3) < \ldots <$ is just a copy of $\mathbb{Z}^+$ order-wise (with minimum $(1,1)$) and we have the same ordering for the points starting with $2$: $(2,1) < (2,2) < (2,3) < \ldots <$ and all points of the form $(1,n)$ are smaller than any point of the form $(2,m)$, so we have two copies of $\mathbb{Z}^+$, one all to the left of the other copy.

The order topology on it is generated by open intervals $((i,n),(j,m))$ with $(i,n) < (j,m)$, plus (for the minimum $(1,1)$) all sets of the form $[(1,1), (i,n))$ as well.

But if $(i,n)$ is in $\{1,2\} \times \mathbb{Z}^+$, and $n \ge 2$ then $\{(i,n)\} = ((i,n-1),(i,n+1))$ and so those singleton sets are open and the order topology is locally at those $(1,n)$ the discrete topology (lots of isolated points).

Also, $\{(1,1)\} = [(1,1), (1,2))$ so the minimum is also an isolated point.

Only $(2,1)$ is interesting: it's not the minimal element of the whole set, so a basic open neighbourhood of it looks like $((1,n), (2,n))$, an open interval with left endpoint smaller than $(2,1)$ so this point can only be of the form $(1,n)$ by the definition of the lexicographic order, and the right hand side we can just assume to be the smallest point above it, $(2,2)$. But this open interval never only contains just $(2,1)$, but always has a whole tail $(1,n+1), (1, n+2), \ldots$ as well.

So indeed at this one point $(2,1)$ the space is shown to be non-discrete, it's the only non-isolated point and the whole set $\{1\} \times \mathbb{Z}^+$ is in fact a sequence converging to it. A space is discrete iff all points are isolated points and in such a space there are no sequences that converge (except those that are eventually constant).

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