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$\mathbb N$ is consist of the basis generated by the set $A_n=\{n,n+1,n+2,n+3....\}$ . Then what properties does it have?

  • Hausdorff: for any natural numbers $x$ and $y$ there are no disjoint open sets to start with...so I think it's not Hausdorff.

  • Connectedness: since all open sets have disjoint intersection so we can't find two disjoint open sets so that union is $\mathbb N$

  • Compactness: Any union of sets if it does not contain $A_1$ then it will not be a open cover for $\mathbb N$ but $A_1 = \mathbb N$ and if I about to take a subcover, I can, by taking $A_1$ i.e. $\mathbb N$ itself but it doesn't seem right somehow.

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Hausdorff: You're basically correct. However: "there are no disjoint open sets to start with" isn't entirely true. The empty set is disjoint with anything. But since you're required to find two neighbourhoods, they can't be empty. If you want to be specific, and $x, y$ are your two points, then any neighbourhood of the smaller of the two points must contain the larger of the two points, and thus have non-empty intersection with any neighbourhood of the larger point.

Connectedness: This is fine, as long as you once again remember to specify "non-empty" open.

Compactness: This is entirely fine. An open cover must contain some neighbourhood of $1$, and there is only one neighbourhood of $1$ and that's $A_1$. But $A_1 = \Bbb N$, so $\{A_1\}$ is a subcover, and it's finite.

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  • $\begingroup$ So it is compact right? We can take $\mathbb N$ itself? $\endgroup$ Commented Feb 1, 2019 at 15:17
  • $\begingroup$ @PranitaGupta $\Bbb N$ itself is the only neighbourhood of $1$, so it must be an element in the covering. So yes. Whether you call it $\Bbb N$ or $A_1$ is entirely up to you. $\endgroup$
    – Arthur
    Commented Feb 1, 2019 at 15:18
  • $\begingroup$ In usual topology R is covered by R itself and for any topology X is covered by X itself ... doesn't it makes every topology compact? $\endgroup$ Commented Feb 1, 2019 at 15:22
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    $\begingroup$ @PranitaGupta No, because with an open cover of, say, $\Bbb R$ in the standard topology, $\Bbb R$ itself doesn't have to be an element in that covering. $\endgroup$
    – Arthur
    Commented Feb 1, 2019 at 15:23
  • $\begingroup$ Or maybe when you start to take any cover you don't have to take necessarily R or X like in this case so you won't be having a finite subcover having R or X. $\endgroup$ Commented Feb 1, 2019 at 15:23
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You're right: the fact that any two non-empty open sets intersect implies both that $\mathbb{N}$ is not Hausdorff and is connected (hyperconnected, this is called). And as the only neighbourhood of $1$ is $A_1=\mathbb{N}$ this set must be in any open cover, making for a one-element subcover. So compact it is.

The definitions tell us, and they are right.

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So the main observation here is that any open subset in the given topology is actually either $A_n$ or $\emptyset$. That's because $A_n$'s are closed under arbitrary unions and intersections (and this is a consequence of the fact that $\mathbb{N}$ is well ordered and I leave it as an exercise).

So any two nonempty open subsets of $\mathbb{N}$ have a nonempty intersection. Therefore $\mathbb{N}$ is not Hausdorff and is connected out-of-the-box.

It is also compact. That's because if $\{A_n\}_{n\in I}$ is a covering then $0$ (or $1$, depending how you count naturals) belongs to some $A_i$. The only choice of such $A_i$ is $A_0$ which is equal to $\mathbb{N}$ and so $\{A_0\}$ is the finite subcover of $\{A_n\}_{n\in I}$.

So this follows because $0$ is special - it is the only element in $\mathbb{N}$ that has only one open neighbourhood - namely $\mathbb{N}$ itself. Any topological space with such a special element is compact.

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