How to think about open sets and continuous functions on discrete metrics I'm working through practice problems for the Math Subject GRE (Which seems to me to be all Analysis and Algebra even though I'd heard it was mostly multivariate calculus). This problem came up:
Let $\mathbb{Z^+}$ be the set of positive integers and let $d$ be a metric on $\mathbb{Z^+}$ be defined as follows:
$d = \begin{cases}
    1,& \text{if } m  \neq n\\
    0,& \text{if } m  =    n              
\end{cases}$
Which of the following is true:
1) For all n $\in \mathbb{Z^+}$,$\{n\}$ is open
2) Every subset of $\mathbb{Z^+}$ is closed
3) Every real valued function defined on $\mathbb{Z^+}$ is continuous.
I have a lot of trouble assessing this claims with a discrete metric and I would like some help clarifying the way I think about them. All three are true, but it's not clear to me why. More specifically:
1) Is every point of such a set an interior point? To assess this, does the open ball we draw have to be (n-1, n+1)? Does it make sense to draw a smaller open ball than this if distances in this metric are either 0 or 1? I feel like the natural language meaning of "interior point" fails here, because the natural idea of "inside" doesn't make sense. 
2) This obviously relies on one, since we test for being closed by examining if the complement is open.
3) It's not clear to me why this is true; part of that is that, as in 1), it's not clear if we can choose epsilons and deltas to be values other than 1 or 0.
 A: Let $X$ be any metric space whose distance function $d$ satisfies
$$d(x,y)=1\;\text{if}\;x \ne y$$
Then for all $x \in X$, the open ball of radius ${\large{\frac{1}{2}}}$, centered at $x$, is just the singleton set $\{x\}$. Hence all singleton sets are open. Note: Although I chose to use radius ${\large{\frac{1}{2}}}$, any positive real $r \le 1$ would have yielded the same singletons sets.

Since every nonempty set is a union of singletons, and arbitrary unions of open sets are open, it follows that all subsets of $X$ are open.

But then all subsets of $X$ are closed (since their complements are open).

Given any function $f$ from $X$ to any topological space $Y$, the inverse image of an open subset of $Y$ is some subset of $X$, hence is open in $X$. Therefore $f$ is continuous.
A: $1.$ Open balls in a metric help you determine just how close two points are.  In a way, the more open balls around one that contain the other, the closer they are.  The discrete metric says all points are equidistant from each other. So, there is a smallest closed ball containing them all and smallest open ball containing only the center.
$3.$ The $\varepsilon$ in the definition of continuity can be any positive real number, including anything between $0$ and $1$.  
A: Regarding 1: Yes every point is an interior point in the topological sense. The radius you use doesn't have to be $1$, just anything less than $1$. The point is that in the discrete topology there is "room" to wiggle around in without bumping into any other points at all. You can see this from the fact that the discrete topology is the subspace topology of embedding $\mathbb{Z}_+$ in, say, $\mathbb{R}$ with the Euclidean topology.
Regarding 3: You can just check it in terms of topology. The point is that there is always a neighborhood around each point where any function is continuous. This is because singletons are neighborhoods, which again is intuitively explained by this "wiggle room" point in the previous paragraph.
A: A set $U$ in a metric space $(X, d)$ is open if for all $x\in U$, there is some $r > 0$ such that $$B(x, r) := \{y\in X : d(x, y) < r\}\subseteq U$$ For $U = \{x\}\subset \mathbb{Z}^+$, $B(x, r) = \{x\} = U$ for all $r\leq 1$, so $\{x\}$ is open in the metric topology induced by $d$ on $\mathbb{Z}^+$ for all $x\in \mathbb{Z}^+$. This implies that the metric topology on $\mathbb{Z}^+$ is discrete. In a discrete topology, all subsets are both open and closed (since every set is a union of points, and unions of open sets are open; every set also has an open complement and is therefore closed), and every function with a domain possessing a discrete topology is continuous (since the preimage of any set will be open, including the preimage of any open set).
