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What's the correct relationship between these two spaces?

I think that topological space is a metric space, since the open is defined by a metric such that $d(x, a) < \epsilon$.

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    $\begingroup$ You're having it backwards. A metric space is a topological space, since the metric induces a topology ("you can define open balls"). But a topological space may not be endowed by a metric ("open sets do not necessarily imply a distance function").. $\endgroup$ – Clement C. Jan 21 '16 at 21:43
  • $\begingroup$ A topological space is a generalisation of a metric space, where you forget about the metric, and just consider the open sets. Every metric space can be viewed as a topological space. But there are topological spaces which cannot be made into metric spaces (for example, the indiscrete topology on any set $X$ with $\#X\ge 2$) $\endgroup$ – Mathmo123 Jan 21 '16 at 21:44
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    $\begingroup$ A more explicit counterexample: let $X$ be a set with at least two points, and consider the indiscrete topology on $X$. It is not a metric space simply because its topology does not separate points. $\endgroup$ – Ángel Valencia Jan 21 '16 at 21:45
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Technically a metric space is not a topological space, and a topological space is not a metric space: a metric space is an ordered pair $\langle X,d\rangle$ such that $d$ is a metric on $X$, and a topological space is an ordered pair $\langle X,\tau\rangle$ such that $\tau$ is a topology on $X$. A metric on $X$ is a special kind of function from $X\times X$ to $\Bbb R$, and a topology on $X$ is a special kind of subset of $\wp(X)$, and obviously these cannot be the same thing. Thus, neither class is technically a subclass of the other.

What is true, however, is that every metric $d$ on a set $X$ generates a topology $\tau_d$ on the set: $\tau_d$ is the topology that has as a base $\{B_d(x,\epsilon):x\in X\text{ and }\epsilon>0\}$, where

$$B_d(x,\epsilon)=\{y\in X:d(x,y)<\epsilon\}\;.$$

Thus, people often say, rather sloppily, that every metric space is a topological space.

On the other hand, it is not true that every topology on a set $X$ can be generated by a metric on $X$. A topological space $\langle X,\tau\rangle$ whose topology can be generated by a metric on $X$ is said to be metrizable. Many, many spaces, even quite nice ones, are not metrizable. Thus, it isn’t true that every topological space ‘is’ a metric space, even in the sloppy sense in which every metric space ‘is’ a topological space.

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  • $\begingroup$ Could you give me an example of a topological space that is not metrizable. $\endgroup$ – ippon Dec 6 '18 at 11:24
  • $\begingroup$ @ippon - Every finite topological space that isn't discrete is not metrizable. $\endgroup$ – Ben Linowitz Jul 6 at 1:22
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Every metric space is a topological space. A subset $S$ of a metric space is open if for every $x\in S$ there exists $\varepsilon>0$ such that the open ball of radius $\varepsilon$ about $x$ is a subset of $S$. One can show that this class of set is closed under finite intersections and under all unions, and the empty set and the whole space are open. Therefore it's a topological space.

Some topological spaces are not metric spaces. Among these are the "long line" (google that in quotes with the additional term "topology" not in the same quotes) and (if I recall correctly) the set of all functions $\mathbb R\to\mathbb R$ with the topology of pointwise convergence.

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Not every topological space is a metric space. However, every metric space is a topological space with the topology being all the open sets of the metric space. That is because the union of an arbitrary collection of open sets in a metric space is open, and trivially, the empty set and the space are both open. Lastly, the intersection of an arbitrary finite collection of open sets in a metric space is also open.

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