Is an open interval in $\mathbb{R}$ a closed set in $\mathbb{R}^2$ under the standard topology of $\mathbb{R}^n$? The standard topology of $\mathbb{R}^n$ is defined:
For any $n\in\mathbb{N}$, let
$$
\mathcal{T}=\{U\subset\mathbb{R}^n|\ \text{if}\ x\in U\ \text{then}\ \exists\epsilon>0\ni B_\epsilon(x)\subset U\}.
$$
In $\mathbb{R}^2$, a 1-dimensional open interval on say the $y$-axis is a subset of $\mathbb{R}^2$, however for a point $y$ in this interval there is no epsilon ball around $y$ which is a subset of the interval since the epsilon ball will contain points in the plane that aren't even on the $y$-axis. I've convinced myself that this is the case according to my understanding of the definitions, but it seems wrong that an open interval in $\mathbb{R}$ would not be closed in $\mathbb{R}^2$.
 A: The set is actually neither open nor closed. (Perhaps these words are a bad choice cause they seem to imply mutual exclusiveness when it is not the case. Sets may be open, closed, both, or neither.)
It is not open for the same reasons you said in your question.
A set is closed if and only if its complement is open. The endpoints of the interval are in the complement of the interval and you can't make a ball around them without including points that are not in the complement. Thus the complement is not open so the interval is not a closed set.
Openness and closedness always depend on the total space. For instance any subset $S\subset \mathbb R^n$ is 'open' when you consider the $S$ as the topological space (and give it the subspace topology) as it must be to satisfy the axiom of topology that says the total space is open. 
A: No, an interval in $\mathbb{R}^2$ is not closed.  Let us consider the "interval" in $\mathbb{R}^2$ $$(a,b)\times\{0\}=\{(x,y)\in\mathbb{R}^2:a<x<b,y=0\}$$
for some $a,b\in\mathbb R$ with $a<b$.  If this were closed in $\mathbb{R}^2$, then its complement would be open.  However, for every $\varepsilon>0$, the ball centered at $(a,0)$ (an element of $\mathbb{R}^2\setminus\left((a,b)\times\{0\}\right)$) intersects $(a,b)\times\{0\}$ nontrivially.  Thus $\mathbb{R}^2\setminus\left((a,b)\times\{0\}\right)$ is not open, and therefore $(a,b)\times\{0\}$ is not closed.

Nevertheless, we can define define a topology on $\mathbb{R}^2$ such that each set of the form $(a,b)\times\{0\}$ is open.  Consider the order on $\mathbb{R}^2$, $\leq$, defined by
$$(\alpha,\gamma)\leq(\beta,\delta)\iff (\alpha<\beta)\ \ \text{or}\ \ (\alpha=\beta\ \ \text{and}\ \gamma\leq \delta)$$ 
for $(\alpha,\gamma),(\beta,\delta)\in\mathbb{R}^2$.
This defines a total order on $\mathbb{R}^2$, hence we can define the order topology with respect to $\leq$ which makes each interval of the form $(a,b)\times\{0\}$ open. 
A: If $I=(a,b)$, $B=I\times \{c\}$ is not closed since $x_n=(a+{{b-a}\over n},c), n\geq 2\in B$ but its limit $(a,c)$ is not in $B$.
