$Bf(a,r) = \{ x \in \mathbb R \mid d(a,x) \le r \}$
I want to show that the the closed ball is contained in the closure of the open ball.
Am guided to take an $x$ from the sphere
Here is what I did,
Let $x$ be an element in $S(a,r)$ then $x$ is in $Bf(a,r)$ because $Bf(a,r) = B(a,r) \cup S(a,r)$.
I then went ahead to show that $x$ is in the closure of the open ball.
Again am guided that I consider a sequence $x_n = x - (1/n)((x-a)/\lVert x-a \rVert)$.
I know that if I take $x_n$ from the open ball $B(a,r)$ the limit of the sequence which is $x$ will be in the closure of the open ball $B(a,r)$.
That would conclude the proof. But then how do I verify that $x_n$ is indeed a sequence of elements in the open ball.
Am working in $\mathbb R$.