# d-metrizable spaces are sequential

By a d-metric on a set $X$ we mean a function $d : X × X \to \mathbb{R}$ satisfying the following two properties:

1. $d(x,y)≥0$ for all $x, y∈X$ and

2. $d(x,x)=0$ for all $x, y∈X.$

The $d$-ball of radius $r>0$ centered at $x$ is defined as $B_r(x)=\{y∈X:d(x,y) < r\}$

We'll say the topological space $(X,T)$ is d-metrizable if there is a d-metric $d$ on $X$ such that $U$ is in $T$ if and only if for all $x \in U$ there exists a $d$-ball centered at $x$ which is contained in $U$. (I.e., the family of $d$-balls is a basis for the topology $T$.)

How can I prove that d-metrizable spaces are sequential?

(A topological space $X$ is sequential means every sequentially open set is open, where $A \subseteq X$ is sequentially open if whenever a sequence $( x_n )_n$ converges to a point in $A$, then $x_n \in A$ for all but finitely many $n$.)

• Are you sure that you dont have the triangle inequality for $d$? – GenericNickname Nov 12 '14 at 8:00
• I don't know, I might. Sure. What would that imply if I do? – user191141 Nov 12 '14 at 8:01
• Not sure yet. Just thought it odd to talk about a metric without the triangle inequality. But I'll think about your problem. – GenericNickname Nov 12 '14 at 8:03
• Please don't ask the same question multiple times. – user642796 Nov 12 '14 at 8:29

Suppose that $U$ is seq. open. Now suppose that $U$ is not in $T$, i.e. $U$ is not open. Then there exists $x \in U$, such that for all $r > 0$ there exists a $x_r \in X$ such that $d(x, x_r) < r$ and $x_r \notin U$. Now let $$y_n := x_{\frac{1}{n}}$$
Then $y_n \to x$ in $X$. Since $x \in U$ and $U$ is seq. open, it follows that $y_n \in U$ for all $n > n_0$ for some $n_0$. This is a contradiction. Hence, $U$ has to be open.