Sequential characterization of open sets in a metric space

Question

Let $(X, d)$ be a metric space. A set $O \subseteq X$ is an open set if for all $x \in O$, there exists a $\varepsilon>0$ such that $B_{\varepsilon}(x) \subseteq O$.

However, I also saw a definition that uses sequences to characterize open sets as follows:

A set $O \subseteq X$ is open if for all $x \in O$ and for all sequences $(x_n) \subseteq X$ that converges $x$, there exists a $n_0$ such that $n \ge n_0$ implies $x_n \in O$.

Can someone explain to me why the second characterization of an open set is equivalent to the first one?

Answer 1

Let's call $(1)$ the open set characterization in terms of $\varepsilon$-neighborhoods and $(2)$ its characterization in terms of sequences.

Theorem. $(1) \iff (2)$

Proof.

Let's prove $(1)\implies (2)$

From the definition of convergence, for all sequences $(x_n)$ converging to $x$, where $x$ is any element of $O$, whatever $\varepsilon$-neighborhood $B_{\varepsilon}(x)$ is chosen there exists an $n_0$ (depending on $B_{\varepsilon}(x)$) such that $\forall n\ge n_0$, $x_n\in B_{\varepsilon}(x)$. But that implies that $x_n\in O$, because from $(1)$ there exists a $B_{\varepsilon}(x)\subseteq O$.

Let's now prove $\mathrm{not} (1) \implies \mathrm{not} (2)$

If $(1)$ is false, that is if for a given $x\in O$ no $\varepsilon$-neighborhood $B_{\varepsilon}(x)$ is included in $O$ then by choosing countably many of them such that $B_{\varepsilon_n}(x)\subseteq B_{\varepsilon_m}(x)$ when $n>m$, a sequence $(x_n)$ with $x_n\in B_{\varepsilon_n}(x)\setminus O$ exists because $B_{\varepsilon_n}(x)\setminus O\ne\emptyset$, $\forall n$, and is convergent in $x$ because for whatever $B_{\varepsilon}(x)$ there exists an $n_0$ such that $B_{\varepsilon_{n_0}}(x)\subseteq B_{\varepsilon}(x)$ from which it follows that $\forall n>n_0$, $x_n\in B_{\varepsilon}(x)$. So a sequence $(x_n)$ convergent to $x$ has been found such that $x_n\notin O$: this imples that $(2)$ is false.

Answer 2

$(1) \Rightarrow (2)$: If $x_n \rightarrow x \in O$, an open set in $X$, then for all $\epsilon > 0$ there is $N \in \mathbb{N}$ with $d(x_n,x) < \epsilon$ for $n > N$. Use this fact with an $\epsilon > 0$ with the property that $B_\epsilon (x) \subset O$.

$(2) \Rightarrow (1)$: If there is some $N$ with $n > N \Rightarrow x_n \in O$, then because of the convergence we also have $d(x_n,x) < \epsilon$ (for some $\epsilon > 0$) for $n > N$. Then for our particular $x$ and $\epsilon$, we have $B_\epsilon (x) \subset O$.

Answer 3

We prove that the sequential condition implies the open ball condition via the contrapositive:

Suppose $E$ is not open. Then $\exists \, x \in E$ such that $\forall \varepsilon >0, \, \, B(x, \varepsilon) \not \subset E.$ Applying this condition repeatedly with $\varepsilon = \dfrac{1}{n}$ we obtain a sequence $(x_n) \in E^c$ such that $d(x_n, x)< \dfrac{1}{n}$ which implies that $x_n \to x$ but $x_n \not \in E$ for each $n \in \mathbb{N}, $ thereby implying that $E$ does not have the stated sequential property.