Let $A = \{\frac{1}{n}:n\in\mathbb{N}\}$. Prove that $f:A\to \mathbb{R}$ is continuous. 
Let $A = \{\frac{1}{n}:n\in\mathbb{N}\}$. Suppose $f:A\to \mathbb{R}$. Prove $f$ is continuous on $A$. 

Definition of continuity: for all $\varepsilon>0$,there exists a $\delta>0$ such that $\lvert x−c\rvert<\delta$ (and $x \in A$) implies $\lvert f(x) − f(c)\rvert < \varepsilon$;
What I did: Fix $c$ in $A$. $\lvert f(x)-f(c) \rvert<\varepsilon$. Then, I do not know what I should choose for $\delta$ such that $x$ will be in $(c-\delta,c+\delta)$  I also do not know the function.
Hence, I think  the definition I used here is not going to work. I should use another one.

Then, I do not know how to use definition 4 to prove it. 
 A: Using the first definition of continuity, given a point $a \in A$ and $\varepsilon > 0$, you want to choose $\delta$ small enough so that all points in $A$ which are within $\delta$ of $a$ have function values within $\varepsilon$ of $f(a)$. As $A$ is discrete, you can choose $\delta$ small enough so that the only point within $\delta$ of $a$ is $a$ itself.
For example, suppose you wanted to show that $f : A \to \mathbb{R}$ is continuous at $\frac{1}{2}$. Let $\varepsilon > 0$. By taking $\delta =  \frac{1}{6}$ (or something smaller), you find that the only the only element of $A$ which satisfies $|a - \frac{1}{2}| < \delta$ is $a = \frac{1}{2}$. But then $\left|f(a) - f\left(\frac{1}{2}\right)\right| < \varepsilon$ is certainly true (because $f(a) = f\left(\frac{1}{2}\right)$).
If you can understand what is going on in the above example, you should be able to write a proof which shows how to guarantee continuity at any point of $A$.
A: The closest elements to $c=1/n$ in $A$ are
$$ \frac{1}{n-1}= \frac{c}{1-c} \quad \text{and} \quad \frac{1}{n+1} = \frac{c}{1+c}. $$
They are distances
$$ c\left( \frac{1}{1-c}-1\right) = \frac{c^2}{1-c} \quad \text{and} \quad c\left( 1-\frac{1}{1+c}\right) = \frac{c^2}{1+c} $$
from $c$. Pick $\delta$ smaller than the minimum of these two, and there is only one element of $A$ in $(c-\delta,c+\delta)$, which is $c$ (and this works for any $\varepsilon>0$).
A: Any function from a discrete topological space to another topological space is continuous, you could expand on that the preimages of open sets on $ \mathbb{R} $ is going to be an open set in $A$.
