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The question is from Topology and Its Applications Chapter 1, by William F. Basner. The question states the following, Let $\mathbb{Z}$ be a topological space with subspace topology inherited from $\mathbb{Z} \subset \mathbb{R}$. Prove that $\mathbb{Z}$ has discrete topology.

Proof. Since $\mathbb{Z} \subset \mathbb{R}$, we note that from the definition of subspace topology we have an $O \subset \mathbb{Z}$ which is only open iff $O' \subset \mathbb{R}$ when $O = \mathbb{Z} \cap O'$. Now, we must precede to show that every $O \subset \mathbb{Z}$ is open, so we can prove that $\mathbb{Z}$ is a discrete space. By definition 6, we see that there must exist a $B_{r}(x) \subset O$ and this must be true for all $O \subset \mathbb{Z}$. Since $O = \mathbb{Z} \cap O'$, $B_{r}(x) \subset \mathbb{Z} \cap O'$ must be true. Therefore, $B_{r}(x) \subset \mathbb{Z}$ and $B_{r}(x) \subset O'$, but because both $\mathbb{Z}$ and $O'$ are open, then $B_{r}(x)$ must be open. Since this is true for all $B_{r}(x)$ we can conclude that $\mathbb{Z}$ has a discrete topology. $\square$

I feel like this proof is a load of crap (It is my first attempt at a proof in a while) I want feed back, and I need to know if I did this correctly.

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It is sufficient to show that every singleton is open in the subspace topology. Since any non-empty set is a union of singletons and arbitrary unions of open sets are open, this will have shown that any set is open in the subspace topology, so that topology is actually the discrete topology.

Let $x\in\mathbb Z$ and let $U\equiv(x-1/2,x+1/2)$. Clearly, $U$ is an open interval and the only integer it contains is $x$. Therefore, $$\{x\}=U\cap\mathbb Z,$$ so that the singleton $\{x\}$ is open in the subspace topology.


My concern with your proof is that while you are using the right definitions, their interpretations are elusive in this context. In particular, the concept of an open ball $B_r(x)$ has a very different interpretation in a discrete metric space than in the usual Euclidean metric space.

For the sake of simplicity, therefore, I would suggest avoiding the metric definition of open sets (in spite of the fact that it is perfectly valid if you are using the discrete metric on the subspace $\mathbb Z$) and stick with the definition that a set in the subspace is open with respect to the subspace topology if it is the intersection of the subspace and an open set in the “main” space.

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Good work, but could be a bit less terse:

Let $\left(\mathbb{R},\tau\right)$ denote the usual topology on $\mathbb{R}$. The subspace topology is given by $\left(\mathbb{Z},\pi\right)$ where $\pi=\left\{ V\cap\mathbb{Z}\right\} _{V\in\tau}$.

Let $W\subset\mathbb{Z}$. Then $V=\bigcup_{w\in W}B_{1}\left(w\right)$ is open in $\left(\mathbb{R},\tau\right)$ and $W=V\cap\mathbb{Z}$, so that $W$ is in $\pi$. Since $W$ was arbitrary, $\pi$ is the discrete topology.

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The proof is a good start, as you've written down the objects you intend to construct and the properties you intend for them to have. That's an important step. The problem is that you haven't gone on to demonstrate the existence of those objects. Given $O$, you have to find a suitable $O'$ and $r$. This is not too hard; try a value of $r$ and see if it works!

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