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I am working through Allen Hatchers Basic Point-Set Topology Notes: (http://www.math.cornell.edu/~hatcher/Top/TopNotes.pdf) and as he introduced the basis of a topology and set remarks on how a basis topology could be done on $R^n$ with open balls. It got me thinking how i can define a simple non metric topology on $R^n$, where i can also find a basis for the topology. (Apart from open disks and open intervals of course).

EDIT: it should not be the trivial Topology with $\{\varnothing,R^n\}$. Sorry forgot to mention that.

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    The trivial topology has basis $\{\varnothing , \mathbb R^n \}$. – 2017-02-06
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    It suffices to find any non separated topology. The one given by Rudy works fine – 2017-02-06
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    What is "a basis topology"? – 2017-02-06
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    i meant basis for a topology. i will fix that. – 2017-02-06

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There are a number of these:

  • The trivial topology.

  • OK, you ruled that out - the "almost trivial topology" $\{\emptyset, \{(0, .., 0)\}, \mathbb{R}^n\}$.

  • In general, any non-$T_0$ topology will do the job (and that's massive overkill).

  • The cofinite topology.

  • The cocountable topology.

And plenty of others. Not only are these not defined in terms of a metric, they don't correspond to any possible metric - they are non-metrizable topologies.

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    $\Bbb R^n$ with one of these topologies is the same as $\Bbb R^m$ with that topology, so they are sort of "boring" in that way. The Zariski topology on $\mathbb R^n$, which has as basis the complements of the zero sets of n-variate polynomials is maybe more interesting in this regard. – 2017-02-06
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You should note that any topology is trivially a basis for itself, so you just need to find an example of a non-metrizable topology (and there are many of these).

However, for a somewhat interesting example that's probably closer to what you had in mind, try a basis consisting of elements of the form $[a_1,b_1)\times\cdots\times[a_n,b_n)$ for $a_i,b_i\in\Bbb Q$.

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    Please Alex, could you add a proof that it is not metrizable? – 2017-02-19