Linking the intuition of topology with its axiomatic definition

Question

For a long time I have had an intuition on what topology is, without ever formally studying it. i.e the classic example of a mug being the same, topologically, as a doughnut, by stretching and reshaping while retaining "holes".

I have recently started studying it formally, with the idea of a topology of a set being closed under unions and finite intersections etc. I can understand these definitions, but in no way can I relate these to my initial intuitive understanding of what topology is; they seem like separate subjects altogether.

Could someone please help bridge this gap in my understanding?

Answer 1

One thing I’ll mention is that the defining properties of a topology (open sets being preserved by arbitrary unions, finite intersections, and the total and empty sets being open) can naturally be derived from a metric space. One way we can approach topology is by looking at these properties as being derived from a metric space (where openness is defined in terms of balls), and to simply drop the notion of a metric.

In a metric space, we have the notion of continuity of functions being defined in terms of an $\epsilon$-$\delta$ definition, but this can be extended to a notion of continuity for topologies. Topology then in a lot of ways is the study of properties that remain invariant under homeomorphisms which are continuous invertible mappings where the inverses are continuous as well. Properties like holes are topological invariants in this sense, but there more sophisticated tools that are needed in order to understand them. These tools include the fundamental group (equivalence classes of loops/closed paths a space will admit) as well as homology/cohomology groups.

Answer 2

The point being made with a coffee and a donut is that the two spaces are homeomorphic, since we can get from one to the other by "reshaping" a space without introducing holes or connecting pieces that weren't connected. In other words, topology is concerned with the question of which spaces are "basically the same" in the loosest possible sense of the "the same": the only kinds of manipulations that are prohibited (in the context of continuous deformation, anyway) are tearing and gluing.

With that being said, let's see how exactly the axioms of topology prohibit "tearing" and "gluing". Consider the following to spaces: $$ X = (-\pi,\pi] \times \{0\} \subset \Bbb R^2, \qquad Y = S^1 = \{(x,y):x^2 + y^2 = 1\} \subset \Bbb R^2. $$ Both $X$ and $Y$ sit in $2$-dimensional space. $X$ is a line segment (which includes its right endpoint but not its left endpoint), and $Y$ is a circle. We could make the line segment into a circle by gluing the ends together, and we can make a circle into a line segment by tearing it apart at one point, but there's no way to get there without breaking one of our rules. So, these transformations should be prohibited, which is to say that we should be able to recognize that the spaces $X$ and $Y$ fail to be homeomorphic. Let's see how the open sets on each space allow us to do this.

By definition, two spaces are homeomorphic if and only if there exists a bijective function $f:X \to Y$ whose inverse $f^{-1}: Y \to X$ is also continuous. Note that in this case, there is a bijective continuous map from $f:X \to Y$, namely $$ f(x,0) = (\cos(x),\sin(x)), $$ but the inverse map from the circle to the segment fails to be continuous. The importance of open sets lies in this notion of continuity: a function $f:X \to Y$ is continuous if and only if for every open set $U \subset Y$, the preimage $f^{-1}(U)$ must be open in $X$.

Note that this definition of continuity generalizes our $\epsilon$-$\delta$ definition (particularly to contexts where there is no sensible notion of distance, i.e. no metric). Consider an $x \in X$. For any margin of error $U$ containing the desired output $f(x)$ in $Y$, there exists a sufficient margin of error $f^{-1}(U)$ around $x$ such that anything "sufficiently close" to $x$ (i.e. within $f^{-1}(U)$) gets mapped to $U$.

Now, let's see why no bijective map from $Y$ to $X$ (which our $f^{-1}$ from above is supposed to be) could be continuous. Note that the open sets on $X$ are given by the unions of open intervals and half-open intervals containing the endpoint. That is, the sets $$ (a,b) \times \{0\}, \quad (a,\pi] \times \{0\} $$ are open for any $-\pi < a < b < \pi$, and so are arbitrary unions. The open sets on the circle are unions of "open arcs". That is, the sets $$ \{(\cos\theta,\sin \theta) : a < \theta < b\} $$ are open for any $a<b$, and so are arbitrary unions. Suppose now, for the purpose of contradiction, that a function $g:Y \to X$ is continuous and bijective. Select a point $(x,y)$ that gets mapped somewhere into the interior of the interval, namely $(-\pi,\pi)$. Since $g$ is a bijection, $g$ maps $Y\setminus \{(x,y)\}$ to $X \setminus \{g(x,y)\}$. Suppose that $g(x,y) = (a,0)$, and consider the sets $$ U = \{(x,0): -\pi < x < a\}, \qquad V = \{(x,0): a < x \leq \pi\}. $$ $U$ and $V$ are connected subets of $X$, and $U \cap V = \emptyset$. By the continuity of $g$, $g^{-1}(U)$ and $g^{-1}(V)$ are open sets in $Y$. Moreover, it must be the case that $g^{-1}(U) \cap g^{-1}(V) = \emptyset$. However, because $U \cup V = X \setminus \{g(x,y)\}$, it must be that $$ g^{-1}(U) \cup g^{-1}(V) = g^{-1}(U \cup V) = Y \setminus \{(x,y)\}. $$ This is a problem because $Y \setminus \{(x,y)\}$ is a connected set. That is, there is no way to break $Y \setminus \{(x,y)\}$ into open sets $A,B$ satisfying $A \cup B = Y$ and $A \cap B = \emptyset$.

This requires proof, but the intuition is clear: either the open arcs comprising $A$ will intersect those comprising $B$, or we will fail to cover the entirety of the remaining arc $Y \setminus \{(x,y)\}$. For a more rigorous proof: $Y \setminus \{(x,y)\}$ is homeomorphic to an open interval in $\Bbb R$, and we can prove that an open interval is connected using one of these proofs.

Answer 3

The idea of a topology perhaps emerges from asking oneself the question: in all of the stretching and reshaping operations that are allowed, what underlying properties of the objects are unaffected? And when the forbidden operations of ripping and tearing occur, what properties of the objects are lost?

Clearly distance itself can be rather severely affected as one stretches, as can other geometric measurements such as angle, straightness, and so on. So much of our traditional geometric intuition goes out the window.

But here's something that doesn't change when we stretch or reshape, and that does change when we rip or tear: closeness.

What is closeness?

Suppose that I take a particular stretching operation, under which a doughnut is stretched and distorted to form a coffee mug. We can think of this operation as a function $f$ from the doughnut to the coffee mug.

If I take two points $x,y$ on the doughnut that are close to each other, their images $f(x)$ and $f(y)$ on the coffee mug will still be somewhat close to each other.

If I take two points $x,y$ on the doughnut that are really close to each other, their images $f(x)$ on the coffee mug will be still closer to each other.

And so on. If you know some analysis, you may see that what I'm describing can be captured in the $\epsilon-\delta$ definition of continuity.

But if I were to actually create a rip, then there are points $x \ne y$ on either side of the rip, which were as close to each other as one could ask for, but after the rip they are not close at all.

So, how can "closeness" itself be described mathematically?

Let's think of "closeness", like beauty, as being in the eye of the beholder. We have an object, a set $X$. We have a bunch of folks observing portions of $X$. The portion of $X$ that Joe observes is a subset of points in $X$, although Joe cannot see everything, and in fact the typical situation is that Joe only sees a small part of $X$. Let's refer to the part of $X$ that Joe sees as his "neighborhood". The key intuition here is that Joe thinks that everything in his neighborhood is "close" to everything else.

The key "closeness" axiom is concocted to resolve potential disagreements between observers over what is close to what. That axiom says:

• If Joe and Jane can see the same point $x \in X$ then there is a third observer, say Charlie, who can also see $x$, and furthermore both Joe and Jane will agree with anything that Charlie thinks is close. To put this in set theoretic terms, Charlie's neighborhood also contains $x$, and Charlie's neighborhood is a subset of both Joe's neighborhood and Jane's neighborhood.

This leads to one of the first definitions in topology:

Given a set $X$, a basis for a topology on $X$ is a collection of subsets $\mathcal B$ of $X$ satisfying the following properties:

1. If $B_1,B_2 \in \mathcal B$, and if $x \in B_1 \cap B_2$, then there exists $B_3 \in \mathcal B$ such that $x \in B_3 \subset B_1 \cap B_2$.
2. For each $x \in X$ there exists $B \in \mathcal B$ such that $x \in B$.

Of course Property 1 is a formalization of the discussion about Joe, Jane and Charlie. Property 2 is there just to ensure that the atlas of neighborhoods includes all the points in $X$.

One last word: Since you already know a bit about the formal definitions of topology, you've probably already seen how the concepts of "topology" and of a "basis for a topology" are related. They are quite close to each other, although they are not identical concepts. I think the concept of "basis for a topology" is more intuitive. It's definitely harder to give an intuitive explanation for the concept of a "topology" itself. Perhaps what's most important in this regard is that the logical structure of the theory becomes simpler when one adopts the concept of a "topology", without any real loss of theoretical power.