interior, closure, and boundary of a graph of a continuous function Suppose $A = \{ (x,f(x)) : x \in \mathbb{R} \} $. $f$ is a continuous function. I want to find the boundary, closure and interior of $A$. We know that $A$ is closed since $f$ is continuous, then we must have that $\overline{A} = A $. However, I'm having hard time trying to find $\operatorname{Int} A$ and $\partial A $. Can someone help me? thanks
 A: Hint: One result you may want to prove (if you don't already know it) is that if $A^\circ$ denotes the interior of $A,$ then $\partial A=\overline A\setminus A^\circ.$ This will make finding the boundary simple, once you've found the interior.
Now, consider the sets $$A_\epsilon:=\bigl\{(x,y)\in\Bbb R^2:f(x)-\epsilon<y<f(x)+\epsilon\bigr\}$$ for $\epsilon>0.$ You should be able to show that $A\subseteq A_\epsilon$ for all $\epsilon>0,$ and that $A=\bigcap_{\epsilon>0}A_\epsilon.$ Now, use this to show that $A$ cannot contain any open ball, and so $A^\circ=$ . . . what?
Added: Take any $(x,y)\in A$--meaning $y=f(x)$--and take any $r>0.$ Note that $(x,f(x)+\frac r2)$ lies in the open ball about $(x,y)$ of radius $r,$ but fails to lie in $A_{\frac r2}.$ Hence, since $A\subseteq A_{\frac r2},$ then $(x,f(x)+\frac r2)$ fails to lie in $A.$ Hence, the open ball about $(x,y)$ of radius $r$ is not contained by $A.$ Thus, $A$ does not contain any open ball (why?), and so $A^\circ=\emptyset.$
A: Let $f: ℝ → ℝ$ be any function. Notice that it does not have to be continuous. Choose any point $p = (x, f(x)) \in \mathrm{Graph}(f) =: M$ and let $U ⊂ ℝ×ℝ$ be an open set around $p$.
Then there is a real number $ε > 0$ such that for the set $R_ε = (x-ε, x+ε) × (f(x)-ε, f(x)+ε)$, we have $p \in R_ε ⊂ U$, because $\{(x - ε, x+ε) | ε > 0\}$ is a neighbourhood basis around $x$ and $\{(f(x)-ε, f(x)+ε | ε>0\}$ is a neighbourhood basis around $f(x)$ in the topology of $ℝ$, making $\{R_ε | ε>0\}$ a neighbourhood basis around $p$ in the (usual) product topology of $ℝ×ℝ$.
Assume that $U ⊂ M$. This implies that $R_ε ⊂ M$ and so the two points $p = (x, f(x))$ and $q = (x, f(x) + \frac{ε}{2})$ from $R_ε$ must lie in $M$.
This can not be the case, because $f$ is a function and thus there is exactly one point in $M = \mathrm{Graph}(f)$ of the form $(x,y)$ – the one with $y=f(x)$.
That means our assumption must be wrong and the converse must be true: $U \not\subset M$.
As $U$ was an arbitrary open set around $p$. The above holds for all open sets $U \ni p$. So $p \notin \mathrm{Int}(M)$.
Likewise $p$ was an arbitrary point of $M$ and so $\mathrm{Int}(\mathrm{Graph}(f)) = M ∩ \mathrm{Int}(M) = ∅$.

So looking from a general topological perspective, as long as the open sets of the target space all contain more than one point, any graph of any function into that space will have empty interior.
A: Try proving $A$ has no interior.
