# Alternative definition for topological spaces?

I have just started reading topology so I am a total beginner but why are topological spaces defined in terms of open sets? I find it hard and unnatural to think about them intuitively. Perhaps the reason is that I can't see them visually. Take groups, for example, are related directly to physical rotations and numbers, thus allowing me to see them at work. Is there a similar analogy or defintion that could allow me to understand topological spaces more intuitively?

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I know this question has been asked here before and also on mathoverflow and I hope somebody can find the relevant posts, but I'd like to mention that metric spaces are much more intuitive and that the axioms of a topological space are abstractions of what happens for metric spaces. –  Grumpy Parsnip Apr 10 '11 at 20:56
to add to the general confusion: the most intuitive definition I know is in non-standard analysis, where topology (or rather uniformity) is replaced by an equivalence relation of "being infinitesimally near" –  user8268 Apr 10 '11 at 20:58
A good approach is to first study metric spaces, which are special cases of topological spaces where the notion of "continuity" is very similar to the notion you'd have learned in calculus. However, it turns out you don't need a notion of distance to get a "topology" on a space, and many metrics can lead to the same "topology" on a set. –  Thomas Andrews Apr 10 '11 at 21:11
In addition, the shear beauty of the "open set" definition is that it makes the definition of continuity of functions between topological spaces nearly trivial. It also turns out that the definition of continuity using opens sets has very interesting and useful meanings when the topology does not come from a metric, for example, when dealing with partial orders. –  Thomas Andrews Apr 10 '11 at 21:46
Related question: math.stackexchange.com/questions/31859/… –  Nate Eldredge Jul 21 '11 at 22:06

There is a MathOverflow question about this very issue; this answer is a nice intuitive explanation, though you will probably also find some of the other answers useful.

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There's also a related question here: math.stackexchange.com/questions/15007/… . –  joriki Apr 10 '11 at 20:59
@joriki, yes, if anything that is a better resource for the OP given their current level. This question should probably be closed as a duplicate then. –  Zev Chonoles Apr 10 '11 at 21:01
It's not an exact duplicate, I think, and your link to MO seems useful, too, so I wouldn't necessarily close this. –  joriki Apr 10 '11 at 21:04

In "Quantales and continuity spaces" Flagg develops the notion of a metric space where the distance function takes values in a value quantale. A value quantale is an abstraction of the properties of the poset $[0,\infty]$ needed for 'doing analysis'. It is then showed that every topological space $X$ is metrizable in the sense that there exists a value quantale $V$ (depending on the topology on $X$) such that the topological space $X$ is given by the open balls determined by a metric structure on $X$ with values in $V$. At this level of abstraction it is thus seen that the open sets axiomatization for topology is nothing but the good old notion of a metric space, only taking values in value quantales other than $[0,\infty]$.

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Think of the half-open interval $(0,1]$ with the usual open sets (e.g. $(1−ε,1]$ is an open neighborhood of $1$.

Then modify the collection of sets considered "open" so that every open neighborhood of $1$ contains some set of the form $(1−ε,1]∪(0,ε)$, i.e. it covers small parts of BOTH ends of the interval. Can you understand that this modification in which sets are considered open also modifies the way in which the space is connected together?

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Sir , You have said "it covers small parts of BOTH ends " can you elaborate a bit more so that i can understand better. I am also suffering from my bad understanding on Top Vector Spaces :( . Thank you. –  Theorem Jun 18 '12 at 13:35
The set $(1-\varepsilon,1]\cup(0,\varepsilon)$ includes a small interval $(1-\varepsilon,1]$ at the right end of the half-open interval $(0,1]$, and also a small interval $(0,\varepsilon)$ at the left end of the half-open interval $(0,1]$. –  Michael Hardy Jun 18 '12 at 14:33