Closure of the real line on $\mathbb R^2$ standard topology

What is the interior and closure of this set A on the standard topology on the real plane?

Let $A= \{(x,0) \mid x \in \mathbb R \}$

Answer I came up with:

$\mathrm{int}(A)= \emptyset$ as there does not exist a basic open set that contains an element in $A$ and contained in $A$.

$\mathrm{Cl}(A)= \mathbb R^2$.

Thanks for the help!

1 Answer

It seems from your comment that you are not using the correct definition of closure. There does indeed exist "a union of closed rectangles that contains the entire real line and is smaller than the entire plane", for example the union of the closed rectangles $\bigcup_{n \in \mathbb Z} ([n,n+1] \times [-r,+r])$, for any choice of $r > 0$. But even knowing that, it's kind of a backwards way of understanding $\text{Cl}(A)$.

In any topological space $X$, for any subset $A \subset X$, the closure of $A$ is the smallest closed subset containing $A$, equivalently the intersection of all closed subsets containing $A$. As a special case, if you happened to know that $A$ was already a closed subset of $X$, then $A = \text{Cl}(A)$.

So let's apply this definition to $X = \mathbb R^2$ and $A$ = $\{(x,0) \mid x \in \mathbb R\}$. The point is that $A$ is closed, because $\mathbb R^2 - A = \{(x,y) \mid y \ne 0\}$ is open (which I'm sure you can verify).

Therefore, $\text{Cl}(A) = A$.