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I cannot use facts involving Hausdorff spaces, as this problem expects knowledge a little bit more elementary.

I am mostly confused with the statement "A topological space (X, $\tau$) is metrizable if there exists a metric d such that $\tau$ is the topology induced by d." The wording here is a little confusing to me. I can vaguely intuitively see that the open balls that construct each U $\in$ $\tau$ must involve the metric d, but as for how this indicates whether or not it "induces" something I am lost at. The definition of a Zariski topology I have been given is:

$\tau$ = {U $\subseteq$ $\mathbb{R}$ : U = $\emptyset$ or U = ($\mathbb{R} \backslash S$) where S is some finite subset of $\mathbb{R}$}

I am thinking to use the method of contradiction here, but without a firm sense of the definitions, I'm having logical difficulties proceeding. Moreover, should I pick an arbitrary set S that fulfills $\tau$'s property?

Lastly, I was confused on how the Zariski topology is even a topology according to this definition. By the properties of a topology $\tau$ on $\mathbb{R}$, we require that $\mathbb{R} \in \tau$. But if for nonempty elements of $\tau$ we have that none contain s $\in$ S, (assuming S is non-empty. My logic here would imply that S must be empty always, so perhaps I am wrong somewhere in this interpretation), then there exists x $\in \mathbb{R}$ such that x $\notin \tau$, demonstrating $\mathbb{R} \nsubseteq \tau$.

Clarification would be much appreciated.

1239asd
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  • $\emptyset$ is finite, so $\mathbb R=\mathbb R\backslash\emptyset$ is in $\tau$. Anyway, elements of $\mathbb R$ aren't supposed to be elements of $\tau$. The elements of $\tau$ are subsets of $\mathbb R$. And consequently, $\mathbb R$ is supposed to be an element of $\tau$, not a subset. – Vercassivelaunos Aug 16 '21 at 12:54
  • How can $\mathbb{R} \in \tau$ when all of $\tau$'s elements are either: $\emptyset$ or $\mathbb{R} \backslash$ S? In other words, doesn't this restrict S to the empty set? – 1239asd Aug 16 '21 at 12:59
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    It's not "choose a specific $S$, and then $\tau$ is ${\emptyset,\mathbb R\backslash S}$". Rather, it's "for every finite set $S$, $\tau$ contains $\mathbb R\backslash S$". – Vercassivelaunos Aug 16 '21 at 13:07

1 Answers1

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Suppose that the Zariski topology on $\Bbb R$ could be induced by a metric $d$. Then, if $r=\frac12d(0,1)$, $B(0,r)$ and $B(1,r)$ would be open sets with empty intersection with $0\in B(0,r)$ and $1\in B(1,r)$.

But there are no such open sets. If $A$ is an open set and $0\in A$, and $B$ is an open set and $1\in B$, then $\Bbb R\setminus A$ and $\Bbb R\setminus B$ are finite sets. Therefore, their union is finite. In other words, $\Bbb R\setminus(A\cap B)$ is a finite set. But therefore $A\cap B\ne\emptyset$, since $\Bbb R\setminus\emptyset$ is infinite.

José Carlos Santos
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  • Can you please explain (1) "there are no such open sets" and (2) "$\mathbb{R} \backslash A$ and $\mathbb{R} \backslash B$ are finite sets?" – 1239asd Aug 16 '21 at 13:13
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    (1) That's what I explain after asserting that there are no such open sets. (2) By definition a subset $A$ of $\Bbb R$ is open with respect to the Zariski topology if and only if $A$ is empty and $\Bbb R\setminus A$ is finite. – José Carlos Santos Aug 16 '21 at 13:22
  • This must've been the source of my confusion here. I didn't interpret the wording of the definition of the Zariski topology properly; I missed the necessary detail that $\mathbb{R} \backslash A$ is finite. Consequently, I wouldn't have thought to reason with this bi-implication. Thank you very much for your help. – 1239asd Aug 16 '21 at 13:54
  • I'm glad I could help. – José Carlos Santos Aug 16 '21 at 14:46