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The definition of a topological space is a set with a collection of subsets (the topology) satisfying various conditions. A metric topology is given as the set of open subsets with respect to the metric. But if I take an arbitrary topology for a metric space, will this set coincide with the metric topology?

I'm trying to justify why we call the elements of a topology "open". If my above question is true, then at least in a metric space, the set of open sets is equivalent to the topology of the metric space. So am I right in thinking that when we remove the metric, we are generalising this equivalence by defining the open sets as those that satisfy the conditions of a topology?

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  • $\begingroup$ I think your question is related to when a topology on a space is metrizable (I.E you can define a metric function that generates the same topology). In general this requires some properties on the topological space itself. $\endgroup$ – DBS Feb 4 '16 at 14:36
  • $\begingroup$ The metrizability depend on the topology ! $\endgroup$ – Surb Feb 4 '16 at 14:37
  • $\begingroup$ What do you mean by "arbitrary topology for a metric space"? Do you mean an arbitrary topology on the points of the metric space? If so, then there is no reason that it will necessarily coincide with the metric topology. The collection of all open subsets in the 'metric topology' will satisfy the general definition for a topology, but there are topologies which are non-metrizable. Thus the notion of a topological space is a generalization of metric space. $\endgroup$ – Apollo Feb 4 '16 at 14:39
  • $\begingroup$ Ok so if the open sets wrt the metric are not equivalent to the open sets in a topology (of a metric space), why are the sets in a topology called open?? I get this is a generalisation but there seems to be no correspondence, despite the same name being used. My problem may be more due to the language more than anything else $\endgroup$ – Ted Jh Feb 4 '16 at 14:48
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Suppose you've got a set, $X$. If you equip $X$ with a metric $d$, now the pair $(X,d)$ is a metric space.

This metric generates a topology on $X$. You consider the collection $\mathcal{B}$ of sets of the form $$ B(x,r):= \{y \in X:d(x,y) < r \}. $$

Now the collection $\mathcal{B}$ is not a topology in and of itself, but lets you build one by taking arbitrary unions and finite intersections of sets from $\mathcal{B}$. The resulting collection $\mathcal{T}$ satisfies the conditions needed for a collection to be considered a topology on $X$. This is exactly the metric topology on $X$ with respect to $d$. So in this setting, a subset of $X$ will be open (is in $\mathcal{T}$) if and only if it is open with respect to the metric (that you can fit a ball of some radius around each point in the set).

But you could just as easily muster up a collection of subsets $\mathcal{T}'$ of $X$ that don't spawn from a metric on $X$, yet still satisfy the conditions to be a topology. The open sets in $(X,d, \mathcal{T})$ will be in general very different from those in $(X, \mathcal{T}')$.

You can build a topology from scratch by declaring, "I want these sets to be open." Then you do what's necessary to make sure everything satisfies the conditions to have a topology. This is the concept of a basis for a topology. So in this setting the idea of "open" in topological spaces is really just an abstraction of the "open" you're used to in metric spaces.

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Except for the trivial case of a metric space with only one element, there is always at least one topology on a metric space that is not the same as the metric topology, namely the discrete topology in which only the empty set and the whole space are open. An example of a very interesting topology on $\Bbb{R}$ that is not the metric topology is the lower limit topology.

As the answer to your first question is "no", you may want to rethink your second question.

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  • $\begingroup$ So in the lower limit topology the "open" sets are the half-open intervals? I just find that a bizarre use of language, probably the cause of my confusion. Why not call it anything else, since it's hardly ever going to correspond to the intuitive idea of open! $\endgroup$ – Ted Jh Feb 4 '16 at 15:10
  • $\begingroup$ I don't know what your intuitive idea of open is, so it's hard to comment on that. $\endgroup$ – Rob Arthan Feb 4 '16 at 20:49
  • $\begingroup$ Just a set that's open wrt a metric. Whereas I wouldn't think to call an arbitrary set that happens to satisfy the topology axioms open. But I'm more happy with the idea now, thanks! $\endgroup$ – Ted Jh Feb 4 '16 at 21:20

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