Does every element of $\mathbb{R}^n$ belong to an open subset on the standard topology? I want to know that for any element $a \in A = \mathbb{R}^n$, there is a open subset of $A$ such that $a \in A$.
Here is my attempt, but I feel like I might be wrong:
Let $a\in A$. Thus, $a$ can be described as $(a_1, ... , a_n)$. I can define an interval $I = (a_1 - r_1, a_1 + r_1) \times ... \times (a_n-r_n, a_n+r_n)$. Because $(a_i-r_i, a_i + r_i)$ is an open interval, $I$ must be an open set. Thus, $a$ is contained in an open set.
I feel like the problem with this is that I'm not proving that $(a_i-r_i, a_i + r_i)$ open $\implies I$ open. Does this go without saying or is it non-trivial? Thanks! 
 A: 
I feel like the problem with this is that I'm not proving that $(a_i-r_i, a_i + r_i)$ open $\implies I$ open. Does this go without saying or is it non-trivial? Thanks! 

That depends on the definition that you use for the topology on $\mathbb{R}^n$. If you are using a metric, then you would have to prove that.
However, given a (finite) cartesian product $X_1 \times \cdots \times X_n$ of topological spaces, the standard topology on the product is that which is generated by the following basis:
$$\mathcal{B} = \{U_1\times \cdots \times U_n \mid U_i \text{ is open in } X_i \}.$$
From this, it immediately follows that $I$ is open. (It is a basis element, in fact!)

Other than that, your proof is correct.

Of course, there is an even simpler solution to your problem as noted in the comments by Chris Custer: $\mathbb{R}^n$ is itself open.

In fact, the above is more general: given any topological space $X$ and $x \in X$, there exists an open set $U$ in $X$ such that $x \in U$. Once again, $U = X$ suffices.
