Prove or disprove that the subspace topology on $\{\frac{1}{n}:n\in\mathbb{N}_+\}\subset \mathbb{R}$ is the discrete topology. Say I have $A = \lbrace \frac{1}{n} | n \in \mathbb{N}_+ \rbrace \subset \mathbb{R}$
I'm trying to figure out if the subspace topology on $A$ that is inherited from $\mathbb{R}$ is the discrete topology.
For the subset $A$ of $\mathbb{R}$, the subspace topology on $A$ is defined by
$\mathcal{T}_A:=\lbrace A \cap U \mid U \in \mathcal{T}\rbrace$ where $\mathcal{T}$ is the standard topology on $\mathbb{R}$.
The largest topology on a topological space $X$ contains all subsets as open sets and is called the discrete topology. In the discrete topology, every point in $X = \mathbb{R}$ is an open set.
In other words, we have to figure out if the subspace topology on $A$ has open set subsets for every point. Disproving the initial statement would be finding a point where this is not true, if one exists.
We know $A\subset \mathbb{R}$. Then the standard topology inherited from $\mathbb{R}$ should be the topology whose open sets are the unions of sets of the type $(a,b)\cap A$, with $a,b\in\mathbb{R}$ and $a<b$.
The issue I'm having is figuring what an open set would look like here. I've been told that some subset $X$ is open in $\mathbb{R}$ if it meets the definition of the topology on $\mathbb{R}$, but $A$ is more than one open interval $(a,b) \subset \mathbb{R}$. Is there a test to apply to $A$ generally?
 A: If it is to be discrete, then each one-point set $\{x\}$ should be an open set in $A$.  Let $x=\frac{1}{n}$ be a point from your set.  Measure the distance to its nearest neighbor:  call that distance $\delta$.  In fact, the nearest neighbor is $\frac{1}{n+1}$, so finding this distance explicitly isn't hard.   Then
$$
\{x\} = A \cap (x-\delta, x+\delta)
$$
and so $\{x\}$ is open in $A$:  we chose $\delta$ so that no one from $A$ (besides $x$) can be within $\delta$ of $x$.
A: As you noted it suffices by the definition of the subspace topology to find for each point $x\in A$ an open neighborhood $U_x \subseteq \Bbb R$ such that no other point $y\in A$ lies inside $U_x$. Then $\{x\}$ is open in the subspace topology and since $x$ was arbitrary $A$ has to be discrete.
Hint: Let $x = \frac{1}{n}$ for some $n>0$. Let $\varepsilon = \frac{1}{2}(\frac{1}n - \frac{1}{n+1})$ and consider $U_x=(x-\varepsilon, x+\varepsilon)\subseteq \Bbb R$. Can you show that $\frac{1}{m} \notin U_x$ for all $m\neq n$?
A: In a $T_1$ first countable space (which is what all metric spaces are) if $x$ is not an isolated point of $X$ there is a sequence $z_n$ of distinct points, all $z_n \neq x$, so that $z_n \to x$. So no $\frac{1}{m}$ in $X$ can be a non-isolated point, or such a $z_n$ would define a subsequence of $(\frac{1}{n})_n$ that converged to $\frac1m$ while it's clear that $\frac1n \to 0$ in $\Bbb R$ and ditto for all its subsequences. This would contradict the unicity of limits of sequences (which holds in all Hausdorff spaces and so in $\Bbb R$).
This is a more "meta" general topology style argument and might be useful to someone. No gory details and more general principles. All convergent sequences in "decent" (Hausdorff is enough) spaces have a unique limit and the other points are isolated in the subspace.
A: Note that open intervals are open in the Euclidean topology on $\mathbb{R}$. Then $\{1\}=(\frac{1}{2},\infty)\cap A$ is open in the subspace topology on $A$, and for $n\geq2$, we have that $$\left\{\frac{1}{n}\right\}=\left(\frac{1}{n+1},\frac{1}{n-1}\right)\cap A$$ is open in the subspace topology on $A$. So the singletons $\{\frac{1}{n}\}$ in $A$ are all open. Since every subset of $A$ is a union of singletons, and therefore a union of open sets, all subsets of $A$ are open in the subspace topology. Since all subsets are open, the subspace topology is equal to the discrete topology on $A$.

To get a better feel for how this works, I shall provide a generalisation to arbitrary topological spaces:

Let $X$ be a topological space with a non-empty subset $A$ and suppose that for each $a\in A$ there exists $U$ open in $X$ with $U\cap A=\{a\}$. Then the subspace topology on $A$ is equal to the discrete topology on $A$.

Proof: For each $a\in A$, we have that there exists $U$ open in $X$ with $U\cap A=\{a\}$. It follows from the definition of the subspace topology that $\{a\}$ is open in $A$. Since each subset of $A$ is a union of singletons, each of which is open, it follows that every subset of $A$ is open in the subspace topology.
