Making Spaces

This is a set of notes from some talk way back in February 2011 and I decided to type it out and it continues naturally from the last post on set theory.

The question posed is what is a space? More often than not, physicists tend to think that spaces are ‘places’ where objects can be contained in very much like the space(-time) we live in. But these ideas have further properties with them such as breaking them into subregions, joining regions into bigger ones, distinguishability of points and regions, all of which are captured in the mathematicians’ notion of topology.

A topology on a set X is a collection \mathcal{T} of subsets of X called open sets satisfying

  1. X and \emptyset are elements in \mathcal{T}
  2. Closure under finite intersections
    If U_1, U_2, \dotsc , U_n \in \mathcal{T}, then U_1\cap U_2 \cap \dotsc U_n \in \mathcal{T}.
  3. Closure under arbitrary unions
    If \{ U_\alpha \}_{\alpha\in A} is any collection of elements in \mathcal{T} (labelled by index set A, then \cup_{\alpha\in A} U_\alpha \in \mathcal{T}.

The pair (X, \mathcal{T}) is called a topological space. Often the case, physicists tend to refer to X as being the space itself with the understanding some standard choice of topology has been made.

Before elaborating on further ideas of topological space, we need to make known some intuitive ideas of what is an open set. First, we take the set \mathbb{R} with an ordered relation <. Intervals of \mathbb{R} form needed subsets. For example,

(a, b) = \{ x\in\mathbb{R} \vert a < x < b\} is an open interval.

[ a, b] = \{ x\in \mathbb{R} \vert a \leq x \leq b \} is a closed interval.

Note that (a, b) form the open sets while [a, b] form the closed sets. The open intervals in fact form the standard topology on \mathbb{R}.

Closed sets are different creatures but not unrelated.  A set A if topological space X is closed if set X-A is open i.e. (X-A)\in \mathcal{T}. For example, [a, \infty) is closed since (-\infty, a) is open. In fact there is some form of duality relationship between the two. One can even define a topological space using closed sets.

Let X be a topological space. Then

  1. Arbirary intersections of closed sets in X are closed.
  2. Finite union of closed sets in X are closed.

Note also the dual behaviour of the set operations of union and intersection. It is important to note that X, \emptyset are both open and closed. Taking the earlier as definition then the empty set is closed since its complement X is open;  and X is also closed since its complement \emptyset is open. A set that is both closed and open is called clopen.

To prove the first point, consider \{ A_\alpha \}_{\alpha\in J} of closed sets. By deMorgan’s law, we have

X - \cap_{\alpha\in J} A_\alpha = \cup_{\alpha\in J} (X - A_\alpha).

But (X-A_\alpha) is open and arbitrary union of open sets are open. Hence (X-\cap_{\alpha\in J} A_\alpha) is open, which then implies \cap_{\alpha\in J} A_\alpha is closed.

For the second point, consider closed sets A_i for i=1,\dotsc , n. By de Morgan’s law, we have

X - \cup_{i=1}^n A_i = \cap_{i=1}^n (X-A_i)

Since (X-A_i) is open and finite intersections of open sets are open, then (X-\cap_{i=1}^n A_i) is open or \cap_{i=1}^n A_i is closed. QED.

A set can also be neither open nor closed. Consider \mathbb{R} with the standard topology. The intervals (-\infty, a) and (a,\infty) are both open but [ a,b) is not open and it is also not closed. This can be understood in the following way; its complement being

\mathbb{R} - [a,b) = (-\infty, a) \cup [b,\infty)

is the union of an open set and a closed set. Thus the said property of [a,b).

It is important to note that the open property of a set is not really universal. We can see this by setting up a counterexample. Consider the set Y = [0,1]\cup [2,3]\subset \mathbb{R} as a topological space with the standard topology. Now [0,1] is open in Y since Y - [0,1] = [2,3] is closed. But then [0,1] is closed in \mathbb{R}; hence the open property does not carry over to the container space.

Having said much about the idea of open sets that underlies the definition of topological space, we can now see the reason why in defining (X,\mathcal{T}), we only allow for finite intersections of open sets. If this limitation is lifted, then one can consider for example, the intersection of all intervals (-\frac{1}{n}, n) of \mathbb{R}. The result would give the set \{ 0 \}, which is not open (hence failing the condition).

Let’s try to go beyond one-dimensional \mathbb{R} and discuss the analogue of open intervals for open sets. One problem is the ordering property of \mathbb{R} is no longer there. Thus one has to invoke other structures for the purpose of introducing open sets. Consider \mathbb{R}^2 as the metric space (\mathbb{R}^2, d) with standard (Euclidean) metric d. Then

D = \{ y\in\mathbb{R}^2 \vert d(x,y) < R\}

is an open disk centred at x\in \mathbb{R}^2 and this serves as the open sets for \mathbb{R}^2. Taking its closure,

\overline{D} = \{ y\in\mathbb{R}^2 \vert d(x,y) \le R\}

is a closed disk and this serves as the closed sets. It is now easy to see how this can be generalised to an open (closed) ball in (\mathbb{R}^n, d) and in higher-dimensional metric spaces. In fact, the generalisation does not stop there. One can build spaces much different from the usual familiar notion of spaces such as space of functions (important in functional analysis), spaces of lines, spaces of solutions etc. For a good read of examples of topological spaces, try this. A route to further generalisation in this case is to move from open sets to related ideas of neighbourhoods of points and to ideas of basis and filiters over which issues like convergence can be discussed. This is beyond the scope of the present post.

Let us however get more elementary for the moment and consider sets of discrete elements to demonstrate further ideas of topology. Consider the following examples over the set X=\{1, 2, 3\}.

Let \mathcal{T}_1 = \{ \emptyset, \{ 1. 2 \}, \{ 2, 3 \}, X\}. Checking on its topology property e.g.

\{ 1, 2\} \cup \{ 2, 3\} = X \in \mathcal{T}_1

which seems to be fine but the following property

\{ 1, 2 \} \cap \{ 2, 3\} = \{ 2\} \notin \mathcal{T}_1

fails. Thus \mathcal{T}_1 is not a topology on X.

If we let \mathcal{T}_2 = \{ \emptyset, \{ 1\}, \{ 2,3\}, X \}, then a check on its topological properties shows that it is a valid topology on X e.g.

\{ 1\} \cap \{ 2, 3\} = \emptyset \in \mathcal{T}_2

\{ 1\} \cup \{ 2, 3\} = X \in \mathcal{T}_2

Next, we can consider \mathcal{T}_3 = \{ \textrm{all subsets of } X\}. It can be easily checked that \mathcal{T}_3 forms a topology on X. The topology is called a discrete topology on X and (X, \mathcal{T}_3) is called a discrete space.

We could also form the topology \mathcal{T}_4 = \{ \emptyset , X \} and this is said to be the trivial topology on X. The different topologies can be depicted in the following picture.

The above example shows that one can define different topologies on a set/space. Note also that \mathcal{T}_4 \subset \mathcal{T}_2 \subset \mathcal{T}_3 and we say this as \mathcal{T}_4 is coarser than \mathcal{T}_2 and \mathcal{T}_3 is finer than \mathcal{T}_2. The finer topologies allow us to distinguish more points of the space (later we will discuss separation axioms).

Now, two different definition of topologies may not necessarily give different topological  spaces. Consider back the metric space (\mathbb{R}^2, d_E) with Euclidean metric  d_E (x,y) = \left(\sum_{i=1}^2 (x_i - y_i)^2\right)^{1/2} whose open set is the open disk. We can equip the space with a different metric, the taxicab metric, d_T (x,y) =\sum_{i=1}^2 \vert x_i - y_i \vert whose open set will be an open square. One can inscribe the square in the disk indicating the d_T-induced topology finer that the d_E-induced topology. But so can the disk be inscribed in the square indicating the reverse. Herce in this case, they are equivalent as topological spaces.

The final point for this post is how can we separate points in space. Often mentioned in theoretical physics literature, when defining a manifold, is the Hausdorff property. A topological space X is Hausdorff if given any pair of distinct points x_1, x_2 \in X, there exists neighbourhoods (open sets cointaining respective points) U_1 of x_1 and U_2 of x_2 such that U_1 \cap U_2 = \emptyset. Note the utility of open sets separating points.

There can be different degrees of variability on how we separate points. We list the different axioms of separability for reader’s reference.

T_0 axiom: If x_1, x_2 \in X, there exists open set O \in\mathcal{T}  such that
(x_1\in O \wedge x_2\notin O) \vee (x_2\in O\wedge x_1\notin O).
A T_0 space is called a Kolmogorov space.

T_1 axiom: If x_1, x_2 \in X, there exist open sets O_1, O_2 \in\mathcal{T} such that
(x_1\in O_1 \wedge x_2\in O_2) \wedge (x_2\notin O_1 \wedge x_1\notin O_2).
A T_1 space is called a Frechet space.

T_2 axiom: If x_1, x_2 \in X, there exist open sets O_1, O_2 \in\mathcal{T} such that
(x_1\in O_1 \wedge x_2 \in O_2) \wedge (O_1\cap O_2 = \emptyset).
This is the case of the Hausdorff space.

T_3 axiom: If A is a closed set with x_2\notin A, there exist open sets O_A, O_2 such that
(A\subset O_A \wedge x_2 \in O_2) \wedge (O_A \cap O_2 = \emptyset).

T_4 axiom: If A, B are disjoint closed sets in X, there exist open sets O_A, O_B such that
(A\subset O_A \wedge B\subset O_B) \wedge (O_A\cap O_B = \emptyset).

T_5 axiom: If A, B are separated sets (i.e. \bar{A} \cap B = \emptyset = A \cap \bar{B}) in X, there exist open sets O_A, O_B such that
(A\subset O_A \wedge B\subset O_B) \wedge (O_A\cap O_B = \emptyset).

Combining conditions T_1 \wedge T_2 gives what is known as regular space and combining T_1 \wedge T_4 gives a normal space. We also have the T_{2\frac{1}{2}} space where the T_2 axiom is supplemented by the condition \bar{O}_1 \cap \bar{O}_2 = \emptyset, which is called a completely Hausdorff space. Most of these differences are however ignored by theoretical physicists.

References

List of books referred to for this post.

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One response to this post.

  1. I just know that there exist a ‘clopen’. Thanks!

    Reply

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