8

Having read through the mathematical definition of endowing a set with a topology I must admit that I'm still struggling to conceptualise what such a mathematical construct is. I've read articles that talk about a topology on a set as defining a notion of "nearness" between elements of that set without needing to introduce the notion of distance (a metric). Is what is meant by this that, say for example I have a set $X=\lbrace a,b,c,d\rbrace$, then I can define a topology on this set $\tau = \lbrace \emptyset ,\lbrace a\rbrace, \lbrace a,b\rbrace, \lbrace a,b,c\rbrace, \lbrace a,b,c,d\rbrace\rbrace$. Now, as $\lbrace a\rbrace\cup\lbrace a,b\rbrace = \lbrace a,b\rbrace$ and $\lbrace a\rbrace\cup\lbrace a,b\rbrace\cup\lbrace a,b,c\rbrace = \lbrace a,b,c\rbrace$, would it be correct to say that $a$ is "nearer" to $b$ than it is to $c$ as $b$ is contained within a smaller neighbourhood of $a$? Is this what is also meant when people speak of two objects being topologically equivalent, as although they may look very different, the relationships between the points remain the same, i.e. $b$ is still "nearer" to $a$ than $c$ even though there may be a huge number of different points added to the neighbourhoods of each of them under some continuous deformation?

Will
  • 3,265

1 Answers1

16

A topology doesn't encode the idea of 'nearness' between points in the same way a metric does. Instead, it tells us when a point is 'near' to a set.

If we restrict ourselves to Hausdorff spaces for simplicity, then two points are either the same, or they are different points, and there is no topological distinction between two points at a distance of $1$ and two points at a distance of $2$. However, it does make sense to say that the point $0$ is 'near' to the open interval $(0,1)$ - it is not contained in that interval, but there are points in $(0,1)$ that get arbitrarily close to $0$.

If $X$ is a topological space, $x\in X$ and $S\subset X$ is any subset, we say $x$ is near to $S$ if $x$ is contained in the closure of $S$ - the intersection of all closed sets containing $S$. For example, the closure of $(0,1)$ in $\mathbb R$ is $[0,1]$, which contains $0$, so we can say that $0$ is near to $(0,1)$.
$\DeclareMathOperator{\cl}{cl}$ We even get an alternative definition of a topology this way. Given a set $X$, a closure operator on $X$ is a function $\cl\colon\mathcal P(X)\to\mathcal P(X)$ such that:

  1. $\cl(\emptyset)=\emptyset$
  2. For all $A\subset X$, $A\subseteq \cl(A)$
  3. For all $A,B\subset X$, $\cl(A\cup B)=\cl(A)\cup\cl(B)$
  4. For all $A\subset X$, $\cl(\cl(A))=\cl(A)$

Think of $\cl$ as being the function that takes a subset $A\subset X$ and returns the set of points of $X$ that are near to $A$. Then the axioms can be stated as:

  1. No point is near to the empty set.
  2. Every point of $A$ is near to $A$.
  3. The points of $X$ that are near to $A\cup B$ are the points that are near either to $A$ or to $B$.
  4. If $x$ is near to the set of points that are near to $A$, then $x$ is near to $A$.

Given a closure operator $\cl$, we can define a topology on $X$ by setting the closed sets to be those sets $A\subset X$ such that $\cl(A)=A$. In particular, by axiom (4), $\cl(A)$ is always a closed set for any $A\subset X$.

Exercise: check that the family $\mathcal K$ of subsets $A\subset X$ satisfying $\cl(A)=A$ is closed under intersections and finite unions; i.e., it is the family of closed sets for a topology on $X$.

Conversely, given a topological space $X$, we can define $\cl$ to be the usual closure operator on $X$; i.e.,

$$ \cl(A)=(\textrm{intersection of all closed sets containing }A) $$

Exercise: Check that $\cl$ satisfies the axioms of a closure operator given above.

The third exercise should convince you that the definition by closures gives us an alternative definition of a topological space:

Exercise: Let $(X,\tau)$ be a topological space, and let $\cl$ be the usual closure operator as defined immediately above. Show that the family $\mathcal K$ induced by $\cl$ is precisely the family of closed sets of $X$.

Conversely, let $(X,\tilde{\cl})$ be a set together with a closure operator $\tilde{\cl}$, and let $\mathcal K$ be the induced family of closed sets. Define a topology $\tau$ on $X$ by $\tau = \{X\setminus L\colon L\in\mathcal K\}$. Show that the usual closure operator on $(X,\tau)$ coincides with $\tilde{\cl}$.

Extension: Recall that a function $f\colon X\to Y$ between topological spaces $(X,\tau),(Y,\sigma)$ is called continuous if $f^{-1}(U)$ is open in $X$ whenever $U$ is open in $Y$ (if you like, $f^{-1}(U)\in\tau$ for all $U\in\sigma$). Give a definition of a continuous map in terms of closure operators.

Further extension: Study what happens if we omit axiom (4) in the definition of a closure operator (and have a chat with me to discuss what you're discovered!)

John Gowers
  • 24,959
  • Thanks for your detailed answer. I'm still unsure as to how, for example, defining a topology on a manifold endows it with a geometrical structure? Also, is what I put about two objects being topologically equivalent heuristically correct at all? Apologies, I didn't word the original post very well, I understand that a topology tells how 'near' points are to a set and not about any kind of distance, but is a topology essentially providing a relation between points and other sets of points? Can one capture the 'nearness' of two points with a topology at all? – Will Jun 12 '15 at 12:29
  • Very nice answer, even me,as a non-mathematican have learned a lot from it! (+1) – tired Jun 12 '15 at 13:08
  • 3
    @Will Good question. Choosing a topology for a manifold does not endow the manifold with a geometric structure, only a topological structure, which is a strictly weaker bit of information. Loosely speaking, if you know what the topology is, then you know what continuous maps are, but you don't know anything about smooth maps or distances between points. – John Gowers Jun 12 '15 at 13:19
  • @Donkey_2009 So is the geometrical structure only introduced by defining a metric on the manifold? Does a topology then just basically provide information on the how a point is related to various subsets of the set? – Will Jun 12 '15 at 13:33
  • @Donkey_2009 From a physics perspective, is the whole point of defining a topology on a manifold to ensure that it can be deconstructed into overlapping open subsets that can be "stitched together" to reconstruct the manifold, and map locally to $\mathbb{R}^{n}$? – Will Jun 12 '15 at 16:42
  • @Will Classically, manifolds were defined to be certain subsets of $\mathbb R^n$ that were locally homeomorphic to $\mathbb R^k$ for some $k,n$. In the more modern definition of an abstract manifold, manifolds are entities in their own right, rather than being subsets of a real vector space. Mathematicians have decided to use topological spaces as the base objects in geometry (though this is not always the case), and these topological spaces are then endowed with a geometric structure (normally in the form of an atlas or a structure sheaf). – John Gowers Jun 12 '15 at 18:59
  • There is such thing as a topological $k$-manifold, which is a topological space that is locally homeomorphic to $\mathbb R^k$. This is a less rigid concept than that of a smooth manifold. It is possible to have two manifolds that are isomorphic as topological manifolds (i.e., they are homeomorphic), but not as smooth manifolds (they are not diffeomorphic). – John Gowers Jun 12 '15 at 19:01