Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Many, infact all the books on topology I have come across define open sets in the following way:

"A set $A$ is said to be open if by moving in small amounts in any direction about any point we land up at a point which belongs to the same set."

Is it so that an open set is always a collection of points only? OR does there exist a general definition of open sets, without taking points into consideration?

share|cite|improve this question
There is something called pointless topology, but I'm not sure it's entirely what you expect. – Henning Makholm Oct 6 '11 at 13:23
@HenningMakholm: Reminds me of Halmos's famous attempt at humor in a Math Review of a paper on a generalization of measure theory, "This paper deals with valueless measures on pointless spaces." – Arturo Magidin Oct 6 '11 at 13:49
Do any books on topology define open sets this way? I'd like to see a list of them. – Byron Schmuland Oct 6 '11 at 13:52
This may be interesting:… – Ragib Zaman Oct 6 '11 at 14:03
The concept of an interior point is pretty useful. – Mark Oct 6 '11 at 18:32
up vote 3 down vote accepted

The "small amounts in any direction" idea doesn't have any direct translation to topology, but another similar idea has an exact definition in topology: here, open sets are intuitively those sets which surround all the points they contain.

The justification of this is as follows:

Start with any topological space and two subsets $A$ and $B$ inside that space. Now in a plain old set, either $A$ and $B$ intersect or they do not. However, in a topological space, we can formalize the idea that $A$ and $B$ 'touch', if not actually intersect. Say that $A$ and $B$ 'touch' if every open set containing $A$ intersects $B$ or every open set containing $B$ intersects $A$ [for future reference: this happens iff 'the closure' of the two sets intersect in the usual sense].

For example, on the real line, $A = [0,1)$ 'touches' $B = [1,2]$. Why? Because any open interval containing $B$ will spill over enough to detect an intersection with the nearby set $A$.

Back to the idea of open sets as surrounding sets. By definition, any point inside an open set $U$ automatically does not 'touch' anything outside that set because by definition the open set $U$ is proof that it doesn't!

This gives a (admittedly rather vague) sense that a point in an open set is spatially separated from the points outside that open set.

share|cite|improve this answer

"moving in small amount" is relevant only to metric spaces.

For a topological space $X$ with topology $\tau$, $A$ is open if $A \in \tau$. Meaning, you define the topology by defining what is an open set in the topology.

For example, if you define a topology $\tau = \{\phi, \{a\}, \{a,b\}\}$ over a finite space $ X = \{a,b\}$, $A$ is open iff $a \in A$ or $A$ is empty.

share|cite|improve this answer
That topology $\tau$ is not the discrete one. – Rasmus Oct 6 '11 at 13:41
Yes, you're right. I'll fix it. – ofer Oct 6 '11 at 13:48
I would just like to point out, we don't have to lose all of our senses and intuition when we generalize open sets to general topological spaces. See the top answer here:… – Ragib Zaman Oct 6 '11 at 14:02
@Ragib: While I agree to some extent with your first statement, I don’t think that sigfpe’s MO answer has much to do with the matter: thinking in those terms is pretty much thinking in metric terms, and that will lead you astray at some point. – Brian M. Scott Oct 6 '11 at 19:37

One could characterize open sets as sets whose points cannot be approached from outside the set. That's probably more of a motivation for a definition than a definition in its own right.

share|cite|improve this answer

If you have a "space" or "ground set" $X$ with points $x$ then any subset $A\subset X$ is a collection of points $x\in X$ $-$ there is no way around that. If some such sets $A$ are declared "open" they necessarily "consist" of points $x$, and if one wants to describe what "openness" for a set $A$ means one is forced to talk about points $x\in A$ and $\notin A$.

In order to get a feeling for "openness" one has to talk about "neighborhoods". It is the very essence of the notion of "topology" on a set $X$ that each point $x\in X$ possesses a system ${\cal V}(x)$ of neighborhoods $V$ of $x$. A neighborhood of $x$ is a preferentially small set $V\subset X$ that contains $x$ and ${\it all}$ points $x'\in X$ which are "sufficiently near" $x$. What "sufficiently near" means is described by axioms (which are obviously fulfilled if nearness is defined by a metric), e.g., if $V$ is a neighborhood of $x$ then $V$ is also a neighborhood of all points $x'$ "near" $x$.

It is essential that ${\cal V}(x)$ contains "arbitrarily small" sets, like balls of radius ${1\over n}$ for arbitrarily large $n$.

Having all this in mind, an arbitrary subset $A\subset X$ is declared open, if $A$ is a neighborhood of each of its points $x$. This means that for each point $x\in A$ the set $A$ contains all points $x'\in X$ that are sufficiently near $x$.

share|cite|improve this answer
At some point one will have to talk about nbhds of points, but one certainly need not define open sets in such terms. I prefer to take topology on $X$ as primitive. – Brian M. Scott Oct 6 '11 at 19:25
@Christian Blatter- Thanks a lot Prof. Blatter. Sir, does it mean that an open set is always a collection of euclidean points ( what we mean by points in euclidean geometry)? – user16186 Oct 9 '11 at 4:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.