Show that if a subset of a metric space is sequentially closed then it is closed

Question

I would like to know if my proof is correct for the following :

Let $(E,d)$ be a metric space and $C\subset E$. If $C$ is sequentially closed then it is closed

My attempt is : Let $C$ be sequentially closed :

$$ \forall x\in C^{\mathbb{N}} : lim_{n\to\infty}x_n = l \implies l\in C $$

Now consider $E\setminus C$, we have : $\forall x\in E\setminus C, d(x,l)>0$ since $l\notin E\setminus C$. Let $x\in E\setminus C$ and consider $\varepsilon = d(x,l)$, we have that $y\in B_{\varepsilon}(x)\implies y\in E\setminus C$. Indeed

$$ y\in B_{\varepsilon}(x)\implies d(x,y)<d(x,l)\leq d(x,y) + d(y,l)\implies 0<d(y,l)\implies y\in E\setminus C $$

Since this is true for all $x\in E\setminus C$, this shows that $E\setminus C$ is open and thus $C$ is closed. Is this seems correct to you ? Thank you a lot.

Answer 1

Let me restate your proof.

1. Let $x\in E\setminus C$ be arbitrary.
2. Define $\epsilon=d(x,l)$.
3. $d(x,y)<d(x,l)\le d(x,y)+d(y,l)$.
4. $0<d(y,l)$.
5. $y\in E\setminus C$.

Step $(5)$ is just wrong; you have asserted it but there is no way to deduce it. The property $d(y,l)>0$ (i.e., $y\neq l$) would not imply that $y\in E\setminus C$ if $C$ has more than one element (say $C$ contains $\ell$ and some other point $p$. Then with $y=p$, $d(y,l)>0$ but it is false that $y\in E\setminus C$).

More pressingly, step $(2)$ is "wrong" in the sense that $\ell$ is undefined. You cannot use an undefined object in a proof. If you want to create some suitable $\ell$, you need to choose a suitable sequence.