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If I am correct, a discrete subset of a topological space is defined to be a subset consisting of isolated points only. This is actually equivalent to that the subspace topology on the subset is discrete topology. There seems no restriction on the cardinality of a discrete subset, i.e. its cardinality can be any.

  1. I was wondering if the following quote from wolfram is true and why?

    Typically, a discrete set is either finite or countably infinite.

    What kinds of topological spaces are "typical"?

  2. Added: Is the following quote from the same link true

    On any reasonable space, a finite set is discrete.

    What kinds of topological spaces does "reasonable" mean?

  3. Is discrete mathematics always under the setting of discrete sets wrt some topologies? In other words, is it a special case of topology theory? Or can it exist without topology?

Thanks and regards!

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For 1, spaces in which the discrete subsets are at most countable, and these are called spaces with "countable spread" in topology. Here, the spread $s(X)$ of a space $X$ is defined as the supremum of the cardinalities of all discrete subspaces of $X$, where by convention a finite supremum is rounded up to $\aleph_0$ (only infinite cardinals are used), also because every infinite Hausdorff space has a countable discrete subset (so spaces with a finite spread would be "pathological" non-Hausdorff spaces, or finite to begin with).

If a space is second countable, then every subspace is second countable too, and a discrete second countable space is at most countable, so a second countable space has countable spread. But this argument can be repeated for other classes of spaces: if every subspace of $X$ is separable ($X$ is then called hereditarily separable) or every subspace of $X$ is Lindelöf ($X$ is then called hereditarily Lindelöf) then $X$ has countable spread too (as a Lindelöf discrete space or separable discrete space both must be countable). For metrizable spaces, countable spread is equivalent to being separable, or Lindelöf, or second countable. See my post on topology atlas, but in general this need not be the case. But the Wolfram quote maybe comes from the fact that a lot of mathematics is done in separable metrizable spaces, like the Euclidean spaces.

An example of a separable compact space that does not have countable spread is $\beta(\omega)$ or $[0,1]^{\omega_1}$.

As to 2, the property that all finite subsets are discrete is equivalent to being $T_1$ (defined either as all singleton sets are closed, or for every $x \neq y$ in $X$, there are open sets $U$ and $V$ such that $x \in U, y \notin U$ and $y \in V, x \notin V$). This already follows from considering subsets of 2 points.

As to 3, adding a discrete topology to a set doesn't make it any more topological, as all functions on it are continuous, there are no non-trivial convergent seuqneces or nets, etc. So a discrete topology adds no information. It's true, for example, that any group can always be given a discrete topology and then it's a topological group (the group operations are continuous), but if we apply theorems from the general theory of topological groups, we cannot prove anything new that we couldn't prove by just plain algebra/group theory. The same holds for other types of (finite or not) structures in discrete mathematics: discrete here is opposite to "continuous", one could say: we do not consider topological or analytical structure, but just the structure as a set. The discrete topology is as informative as no topology in this case....

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Yes. Put the discrete topology on [0,1]. This "typically" means in a separable space.

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Thanks! What does "this 'typically' means in a separable space" mean? – Tim Feb 4 '12 at 2:53
$\ell^\infty$ is not separable. I do not consider it pathological. – ncmathsadist Feb 4 '12 at 3:00
Do you mean ℓ∞ is not typical, but not pathological? – Tim Feb 4 '12 at 3:02
It is, in way, typical. It is actually a Banach Space that is not very "far out." – ncmathsadist Feb 4 '12 at 3:05
Thanks! From the same link, I was wondering why "the set of integers is discrete on the real line"? For each integer, any of its neighbourhood has a real number other than the integer, so the integer is not an isolated point. Thus the set of integers cannot be discrete in the set of reals. Isn't it? – Tim Feb 4 '12 at 3:32

For the added Question $2$, here is an unreasonable space. Underlying set: the reals, or the integers from $1$ to $10$, or indeed any non-empty set. Open sets: the empty set and the whole space, that's all! This is a topology, usually called the indiscrete topology, or the trivial topology.

If $W$ is such a space with more than $1$ element with the trivial topology, then no non-empty subset of $W$ is discrete.

There are quite less extreme examples. One surprisingly important one (it has some applications in Theoretical Computer Science) is the Sierpinski Space. It only has two elements, say $0$ and $1$. The open sets are everything except $\{0\}$. The finite set $\{0,1\}$ is not discrete, since any open neighbourhood of $0$ is the whole space.

As to Question $3$, almost all of discrete mathematics is unconnected with general topology. Algebraic topology is another matter, here there are deep and fruitful connections.

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Thanks! For discrete mathematics, I meant to ask if there is always some (discrete) topology involved? Even some (discrete) topology not explicitly stated. – Tim Feb 4 '12 at 3:24
@Tim: Saying no, there generally isn't sounds absolute, but it is accurate. If we are working on structured finite sets, such as graphs, sure we can put the discrete topology on the vertices, but that basically means we won't use topology, since then notions like continuity can tell us nothing, everything is continuous. – André Nicolas Feb 4 '12 at 3:32
@AndréNicolas a set of one point is always discrete, because the only topology on a singleton is discrete (= indiscrete, for this case). But for indiscrete spaces all 2 point sets are non-discrete, which characterizes indiscrete spaces. These spaces can be called "nowhere discrete" in a way. – Henno Brandsma Feb 4 '12 at 9:30
@AndréNicolas in the Sierpiński space (with $0$ isolated) both singletons are discrete in your definition (which is the standard one and coincides with mine) as for $A = \{0\}$ we pick $N = \{0\}$ and for $A = \{1\}$ we pick $N = \{0, 1\}$; in fact for any singleton we can always pick $N = X$ to show the singleton set is discrete. There is discrete (in itself), which is your definition, and the definition that $A$ has no limit points (a.k.a. closed and discrete), which makes $\{0\}$ non-discrete in that definition. But $\{0\}$ is discrete but has $1$ as a limit points, so is not closed. – Henno Brandsma Feb 4 '12 at 10:02
@AndréNicolas the second definition comes down to "every $x \in X$ has a neighbourhood $N$ such that $N \cap A \subset \{x\}$" and this just says that "no $x$ is a limit point of $A$", or "$A' = \emptyset$", or "$A$ is closed and has the discrete topology (as a subspace)". – Henno Brandsma Feb 4 '12 at 10:06

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