General Topology: Neighborhood Show that a subset $U$ of the real numbers is open in the usual topology if and only if, for all $x$ in $U$, there is a number $\epsilon>0$ such that $|y-x|<\epsilon$ implies $y$ is in $U$.
"I don't know what to try and prove here, it seems that the implication where we suppose $U$ is open in the usual topology follows from definition of how we measure distance in n dimensional Euclidean space." 
I really want to learn this stuff, so I would just like some information maybe making the problem simpler, or an example if possible. thanks.
 A: First, the open sets of standard/usual topology on $\mathbb{R}$ is defined by the $\epsilon$-balls (These are the basis elements of the standard topology - and also their unions)
$B_{\epsilon}(x)= \{y \in \mathbb{R}\ |\ d(x,y)< \epsilon \}$
Where $d(x,y)$ is the Euclidean metric for the reals. ($d(x,y)=|y-x|$)
($\Rightarrow$) Suppose $U$ is open in the standard topology for the reals. Suppose further that for all $x \in U$, there exists an $\epsilon>0$ such that $|y-x|<\epsilon$.
We want to show that $y \in U$.
Note that $U$ is a union of some of the $\epsilon$-balls mentioned above, by the definition of the standard topology on $\mathbb{R}$.
Let $x \in U$. Now since there exists an $\epsilon>0$ such that $d(x,y)=|y-x|<\epsilon$, we have that 
$$y \in B_{\epsilon}(x)$$
by definition of the $\epsilon$-balls.
And since $U$ is a union of these balls, we have that
$$y \in U$$
This concludes the first part of the proof.
($\Leftarrow$) Suppose that for all $x \in U$, there exists an $\epsilon>0$ such that $|y-x|<\epsilon$ implies $y \in U$.
We will use this definition (it's easier!) to prove it instead of the "union of basis elements" definition. (You can show that the definition below is equivalent to that of the "union of basis elements" definition)
Definition: Given a set $X$ with a basis $\beta$, we say that a set $A \subset X$ is open if for each $x \in A$, there is a $B \in \beta$ such that $x \in B \subset A$. 
Can you see that the supposition is just restating this definition?
A: The goal is to show that statement $A$ ($U$ is open in the usual topology) is equivalent to statement $B$ (every point of $U$ is an interior point; that is, for each $x \in U$ there is a ball around $x$ completely contained in $U$). Everything I have said so far is just a rephrasing of the statement of the above question.
I'm assuming that "usual topology on $\mathbb{R}$" is the one whose basis consists of the open intervals; that is, every open set is either a union or a finite intersection of open intervals.
To show $A \implies B$, you need to show that $B$ holds for each open set we just described. To show $B \implies A$, you need to show that the only subsets of $\mathbb{R}$ that satisfy $B$ are the open sets described above.
A: EDIT: I'll be assuming that open sets are those which can be written as the union of open balls.
$\Rightarrow$: You have to show that for every $x \in U$ there exists an open ball centered in $x$ and contained in $U$. Suppose $x \in U$; then, since $U$ is the union of open balls, $x$ is contained in some open ball $B_{\epsilon}(y)$ of center $y$ and radius $\epsilon$. Let $d := d(x, y)$, then $B_{\epsilon -d}(x) \subset B_{\epsilon}(y) \subset U$.
$\Leftarrow$: If for every $x \in U$ there exists $\epsilon > 0$ such that $B_{\epsilon}(x)$  is contained in $U$, then, since $B_{\epsilon}(x)$ is an open set, it follows from your other question that $U$ is open.
