A manifold $M$ of dimension n is a topological space with the following properties:
a) $M$ is Hausdorff
b)$M$ is locally Euclidean of dimension n
c) $M$ has a countable basis of open sets.

Why is the first property necessary? I do not have much experience with Hausdorff spaces, hence I am not able to see the importance of that condition, probably it is something obvious.

Since I come from a physics background, I can understand the importance of the second property. But what is the strong mathematical motivation to study such special subset of a topological space? Also the second property can be alternatively stated as: each point $p\in M$ has a neighborhood $U$ which is homeomorphic to an open subset $U$ of $\mathbb R^n$. Why should it be homeomorphic to an open set and not closed?

Similarly in the third property,why a countable basis of open sets?

  • 1
    $\begingroup$ Certainly not all mathematicians require their manifolds to satisfy (a) and (c), but on the other hand most of manifolds you're going to write down will have these properties. In any event, one wants to do calculus on manifolds, and to me it's somewhat hard to imagine doing calculus in a non-Hausdorff space. There are probably a few common technical lemmas that are useful in the study of manifolds that rely on this condition (partitions of unity and so on); I'll try to list a few later. $\endgroup$ Mar 17, 2012 at 17:37
  • $\begingroup$ Related question. $\endgroup$ Mar 17, 2012 at 20:27
  • 5
    $\begingroup$ Manifolds are like pornography, hard to define, but you know it when you see it. $\endgroup$
    – galois
    Mar 9, 2015 at 13:19

2 Answers 2


A lot of the work on smooth manifolds is to let us use Euclidean analysis to merely locally Euclidean things that come up. Things that aren't Hausdorff are terrible and scary, real intuition busters (at least in my case), so I don't mind at all that we require that. And in fact, with just these 3 requirements (and a smoothness requirement), smooth manifolds can actually be viewed more or less in real space (the Whitney Embedding Theorem). And our real space intuition is pretty good, you know? And it suffices, has really nice properties, provides some pretty rich material, etc.

But there are some people who care a lot about non-Hausdorff manifolds. In fact, secretly, we see them and don't know it. The etale space of the sheaf of continuous real functions over a regular manifold is a manifold, and it's sometimes non-Hausdorff.

Similarly, there are people who care about non-second-countable manifolds. But these are also a bit unfortunate. One of the great things about second countability is that it guarantees that ordinary manifolds are paracompact. A paracompact smooth manifold admits partitions of unity subordinate to a refinement of any cover. Why is this important? (for that matter, what does it really mean?) While it's easy to stitch together continuous functions to make a continuous function (just sort of join the ends together, right?), it's really hard to stitch smooth functions together in general (join the ends together, and perturb it so the first derivatives align, and so the second align, etc.). But this can be done with little fuss with partitions of unity, and thus with little fuss with second-countability.

And if you study smooth manifolds, you'll see that partitions of unity are immediately used for everything.

So it's the way it is because it has these really nice properties, right? Well, why don't we just require manifolds to be paracompact? (Firstly, there is a distinction, but it's 'small.' A manifold with more than countably many disconnected components may be metrizable, and thus paracompact, but obviously won't be second-countable). In this case, the category of paracompact manifolds is closed coproducts, which doesn't hold for second-countable manifolds. In fact, some people do only consider paracompact manifolds.

At the end of the day, Hausdorff and second countability are exactly what let us use the embedding theorem to view manifolds in real space, and that's what's deemed important for people on their first tour through manifolds.

  • 1
    $\begingroup$ Thanks for the enlightening answer; I often enjoy your writings. $\endgroup$
    – 000
    Mar 17, 2012 at 20:38
  • $\begingroup$ Thank you for your kind reply. It took me some while to understand what you said. Yet I havent completely understood the partitions of unity part, but I am sure with more maturity in the subject I will get it. $\endgroup$
    – Omkar
    Mar 22, 2012 at 6:24

Let us first review our topological notions to ensure there isn't any confusion there.

A topological space is a set $X$ together with a collection $U$ of subsets of $X$, called a topology on $X$, so that the following hold,

  • The empty set $\varnothing$ and the set $X$ are both open.
  • $U$ is closed under arbitrary union: $$\text{If } U_{1},...,U_{k} \text{ are open, then } \bigcup_{i =1}^k U_{i} \text{ is open.}$$
  • $U$ is closed under finite intersections: $$\text{If } U_{1},...,U_{k} \text{ are open, then } \bigcap_{i=1}^k U_{i} \text{ is open.}$$

We can define maps between topological spaces $X$ and $Y$ where, A map $f: X \rightarrow Y$ is continuous if $f^{-1}(U)$ is open for every open set $U \subset Y$. A continuous map that has a continuous inverse is a homeomorphism and if there exists a homeomorphism between $X$ and $Y$, then $X$ and $Y$ are considered topologically equivalent.

Generally, a topological manifold can be thought of as some space which locally looks like $\mathbb{R}^n$ endowed with a metric by which calculus can be performed. More precisely, a topological space $\mathcal{M}$ is locally Euclidean of dimension $n$ if and only if every point of $\mathcal{M}$ has a neighbourhood that is homeomorphic to an open subset of $\mathbb{R}^n$. We then have our initial definition of a topological manifold as follows,

A topological space $\mathcal{M}$ is called a topological $n$-manifold when, $\mathcal{M}$ is locally Euclidean of dimension $n$; that is, for each $p \in \mathcal{M}$ there exists,

  • An open set $U \subset \mathcal{M}$ with $p \in U$,
  • An open set $\widetilde{U} \subset \mathbb{R}^d$ such that $\varphi : U \rightarrow \widetilde{U}$ is a homeomorphism.

Furthermore, $\mathcal{M}$ is a Hausdorff space: For every two distinct points $p,q \in \mathcal{M}$, there are disjoint open subsets $U,V \subset \mathcal{M}$ such that $p \in U$ and $q \in V$. An additional property that $\mathcal{M}$ is second countable (there is a countable basis for the topology of $\mathcal{M}$) gives a more general definition

Examples of topological manifolds, are $\mathbb{R}^n$, the unit sphere $\mathbb{S}^n$, graphs of continuous functions $f: U \rightarrow \mathbb{R}^d=n$, and the real projective plane $\mathbb{RP}^n$.

In general the reason for defining a topological $n$-manifold with the additional characteristics to locally Euclidean is to avoid certain pathological examples that we do not want to count as a manifold. For example, to explain why we would want a space to be Hausdorff, I'll present the best illustration I have found so far: we want all points to be, in a sense, separable. Consider the real line $\mathbb{R}$ glued to another real line $\mathbb{R}'$ along the open sets $(-\infty,0) \cup (0,\infty)$ away from the origin. Then we cannot embed this space in Euclidean space because there are multiple ways to fill in a jump discontinuity of a function, and compact sets are not necessarily closed.

For a space to be second-countable we need to avoid the uncountable well-ordered sets guaranteed to exist by Zorn's Lemma. Firstly recall that if $(X,<)$ is a partial order, then $C \subseteq X$ is a chain in $X$ if $C$ is linearly order by $<$.

That is, Zorn's Lemma states,

If $(X,<)$ is a partial order such that for every chain $C \subseteq X$ there is $x \in X$ such that $c \leq x$ for all $c \in C$, then there is $y \in X$ such that the is no $z \in X$ with $z > x$. In other words, if every chain has an upper bound, then there is a maximal element of $X$.

From this we can give a proof of the well-ordering principle which states any there is always a well-ordering of $A$ (for any nonempty $C \subseteq A$, there is an $a \in C$ such that $a \leq b$ for all $b \in C$). Now, the actual reason we need to have our manifold be second countable, is that Zorn's lemma guarantees the existence on an uncountable well-ordered set. So, given the order topology we can take half-open intervals indexed by our uncountable well-ordered set and have that the space cannot be embedded in $\mathbb{R}^n$ and that there is no countable dense subset (not separable). So, we require our space to be second countable because we want to avoid such pathological examples to do with bizzare uncountable orderings that lose analytic separability.

I will stop rambling now, but hopefully this is helpful; the general idea that I would take away is there there are good reasons for defining a topological $n$-manifold the way it is commonly defined because there are certain "counterexamples" or pathological examples that do not have the nice properties. If we are to classify such a vast collection of spaces, as "all topological manifolds of any dimension", we certainly what the power of generality to not have to deal with pathological examples that cause general theorems to not hold because of some bizzare uncountable ordering or the like.


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