I'm trying to show that the metric topology is indeed a topology.

To do so, I want to show the following three statements are true:

  1. $\emptyset$ and X are in $\tau$
  2. finite intersections of open sets are open
  3. arbitrary (finite, countable, uncountable ...) unions of open sets are open

I'm having trouble with part 1, that is, with what I think I need to do in order to show that $X\in\tau$.

How can I show that $\tau$ contains an open ball $B_{\epsilon}$(x) for every $x\in X$?

  • 2
    $\begingroup$ The metric topology is usually defined as being the topology that has open balls as a basis. For every $x\in X$, take the open ball of radius $1$. Then $X$ is the union of these. $\endgroup$ Jun 7, 2014 at 0:50
  • $\begingroup$ Thanks for responding so fast. Why can we say that radius 1 works? $\endgroup$
    – MerryMC
    Jun 7, 2014 at 0:52
  • $\begingroup$ Sure, the ball contains $x$. $\endgroup$ Jun 7, 2014 at 0:53
  • $\begingroup$ Right, but for instance, what if x is on the boundary of a subset of the real line, maybe Y=[1,4]? The ball of radius 1 around 4 isn't contained within Y. Does that mean that [1,4] with the usual metric can't induce the metric topology? Maybe it doesn't matter since the points beyond 4 are not in the set [1,4], and so are not even in the universe that we care about and thus are irrelevant $\endgroup$
    – MerryMC
    Jun 7, 2014 at 0:59
  • 1
    $\begingroup$ No problem. If we are working with the set $X=[1,4]$, there is nothing else in the world. The open ball of radius $1$ and centre $3.8$ will be the interval $(2.8,4]$. $\endgroup$ Jun 7, 2014 at 1:03

3 Answers 3


You start with a metric $d$ on a set $X$, without a topology.

You then define a topology $\mathcal{T}$ on $X$ by $$O \in \mathcal{T} \text{ iff } \forall x \in O: \exists r>0: B(x,r) \subset O\text{.}$$

Of course, one has to check that this fulfils the axioms for a topology.

To see that $X$ (what you asked about) is in $\mathcal{T}$: for any $x \in X$, we can pick any $r >0$ that we like (so e.g. $1$ will do, but $r=100$ is also OK) and the $B(x,r)$ is always a subset of $X$ (by definition, as $B(x,r) = \{ y \in X: d(x,y) < r \} \subset X$) so the condition is satisfied for any $r$.

The empty set if a bit more subtle: the condition is true vacuously (any statement that starts with $\forall x \in \emptyset$ is true, as there is no $x$ in the empty set to falsify the rest of the statement), so $\emptyset \in \mathcal{T}$.

The definition I gave is equivalent to the condition that a set is open iff it is a union of some family of open balls, which is also a common way to formulate this. Or, if the notion of a base was introduced, we take the set of open balls of $d$ as the base for a new topology, and then we check that the set of $d$-open balls satisfies the axioms for a base for a topology instead.


The induced topology is usually defined as the topology that has the open balls as a base. For every $x\in X$, let $B_x$ be the open ball of radius $1$ and "center" $x$, that is, the set of all points $y$ such that $d(x,y)\lt 1$. Then $X$ is the union of the $B_x$, since $x\in B_x$.


Disclaimer: This post is meant informative. Where necessary things are left out for clarity.

Most varieties of proves rely on the approach given by Hausdorff as follows.

The collection of open balls with center at some chosen point is a filter base, that is: $$\forall B_r(x),B_{r'}(x)\exists B_{r_0}(x): B_{r0}(x)\subseteq B_{r}(x)\cap B_{r'}(x)$$ (That will be our neighborhood base later: $\mathcal{B}(x)$)

The upper set induced by the filter base is a filter then: $$\mathcal{N}(x):=\uparrow\mathcal{B}(x)$$ (That will be our neighborhood filter later: $\mathcal{N}(x)$)

What remains for this to actually deserve the attribute neighborhood is the correlation to its chosen point and the interrelation between points.

Let me first state this in general: $$\forall N\in\mathcal{N}(x):\qquad x\in N$$ $$\forall N\in\mathcal{N}(x)\exists U\in\mathcal{N}(x):\qquad N\in\mathcal{N}(u),u\in U$$ In the case of metric balls that is: $$\forall N: x\in B_r(x)\subseteq N$$ $$\forall N: B_{r(u)}(u)\subseteq B_r(x)\subseteq N$$ (The latter then can be chosen to be the open neighborhoods: $U$)

All together is called the neighborhood system: $$x\mapsto\mathcal{N}(x)$$

There is more to read in the neighborhood attribute and in the definition via neighborhoods.


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