Let $(X, d)$ be a metric space. Prove that for every $r > 0$ and $x\in X$ that $\overline{\rm S_r(x)} \subseteq B_r(x)$

Question

I have a proof for this theorem but I am not feeling great about it. I would love some pointers and help to guide me in the right direction. Here's what I know as starting information:

$S_r(x)$ $=$ {$y\in X: d(x,y) < r$} $=$ the open ball of radius r

$B_r(x)$ $=$ {$y\in X: d(x,y) \leq r$} $=$ the closed ball of radius r

So consequently, it's trivial that $S_r(x) \subset B_r(x)$.

If we let $Y$ be a subset of a metric space $(X,d)$, the closure of $Y$ is the set $\overline{\rm Y} =$ {$x\in X: x\in Y$ or $x$ is a limit point of a sequence in $Y$}

Let $A$ be the set of limit points of $S_r(x)$.

So this means that

$\overline{\rm S_r(x)} = S_r(x)\cup A$

Now for the proof (the bit I need some assistance with).

Proof:

Take a metric space $(X,d)$. We wish to show that for every $r > 0$ and $x\in X$ that $\overline{\rm S_r(x)} \subseteq B_r(x)$. In order to show this, we must show that $S_r(x)\cup A \subseteq B_r(x)$. It is trivial that $S_r(x)$ is a subset of $B_r(x)$. So now we just have to show that $A$ is a subset of $B_r(x)$ because if this turns out to be the case then the union of $S_r(x)$ and $A$ will be a subset of $B_r(x)$. Pick an arbitrary limit point $z \in A$. We know $\exists (Z_n)_{n=1}^{\infty} \subset S_r(x)$ such that $(Z_n)_{n=1}^{\infty} \to z$. We want to show that $d(x,z) \leq r$. Now assume the contrary, that is, $d(x,z) > r$ and take some $y \in S_r(x)$.

Consider the following: $d(x,z) > d(x,y) + d(y,z)$. If this were true then $(X,d)$ wouldn't be a metric space, hence, $d(x,z) \leq r$ and as a result of this $A \subset B_r(x)$ which is what we wanted to show.

$Q.E.D$

Any help would be greatly appreciated, thank you!

Answer 1

You're last paragraph isn't correct. $d(x,z) > d(x,y) + d(y,z)$ would have to be true for every $x,y,z \in X$, and this is not implied by what you've written previously.

Try something more like this: If $y \in A$ ($A$ being the set of limit points of $S$), then there exists a sequence of points $y_n \in X$, such that $y_n \rightarrow y$, i.e. Let $\epsilon > 0$ be given, then $\exists N > 0$ such that $\forall n \geq N$, $d(y_n, y) < \epsilon.$ Then $\forall n \geq N$, $d(x,y) \leq d(x,y_n) + d(y_n, y) < r + \epsilon$. Since $\epsilon$ is arbitrary, $d(x,y) \leq r$.