One of my favorite ideas in all of mathematics is to study the topology of a space by studying functions on the space. This is the underlying idea of Morse theory, which I hope to learn more about. A huge set of examples of this idea that I am more familiar with comes from complex analysis.
One may observe that in the examples that I have given, the functions we are studying to probe the topology possess some non-topological properties. For instance, holomorphic functions famously have some extremely strong properties most of which are not topological in nature at all. The same goes for harmonic functions, which share many properties with holomorphic functions. In general, differentiability is not a property of a function that interacts much at all with the domain topology. So it is natural to wonder if we can study the topology of a space by studying functions that obey no assumption other than the assumption that they interact somehow with the topology. It also seems reasonable that the topology should be uniquely determined by such functions. More precisely, let \(X\) be a topological space and let \(C(X)\) be the ring of real-valued continuous functions on \(X\). Question: Can one recover the topology on \(X\) given \(C(X)\)? In this blog post, we will show that the answer is in the affirmative. We will focus on the following special case: fix \(X\) to be a compact Hausdorff topological space. Consider the spectrum of the ring, \(C(X)\), which we denote \(\text{Spec }C(X)\). We can interpret the spectrum as a topological space by giving it the Zariski topology. Let \(\mathscr{M}\) be the set of maximal ideals of \(C(X)\). Since every maximal ideal is prime, \(\mathscr{M}\subseteq\text{Spec }C(X)\) and we can endow \(\mathscr{M}\) with the subspace topology. We sometimes refer to \(\mathscr{M}\) as the maximal spectrum of \(C(X)\). The incredible fact which we will prove is that \(X\) is homeomorphic to \(\mathscr{M}\). For every \(x\in X\) define \(I_x=\left\{f\in C(X)\colon f(x)=0\right\}\). Clearly, \(I_x\) is an ideal of \(C(X)\). What is less clear is that \(I_x\) is always a maximal ideal. We will show this in two different ways. In the first method, we will show that any ideal properly containing \(I_x\) is the full ring \(C(X)\). Fix \(x\in X\) and pick \(g\in C(X)\setminus I_x\). Since \(X\) is Hausdorff, \(\{x\}\) is closed, and since \(g\) is continuous, \(g^{-1}(\{0\})\) is closed (and disjoint with \(\{x\}\) since \(g\notin I_x\)). Recall that every compact Hausdorff space is normal (\(T_4\)), so by Urysohn's lemma, there exists a continuous function \(f\colon X\to\mathbb{R}\) such that \(f(x)=0\) but \(f(y)=1\) for all \(y\in g^{-1}(\{0\})\). Notice that \(f\in I_x\). Moreover, \(f\) and \(g\) have no common zeros by construction. Therefore, \(f^2+g^2\in\langle f,g\rangle\) is always positive, so the multiplicative inverse \(\frac{1}{f^2+g^2}\) exists in \(C(X)\). Since ideals are closed under multiplication from any element, \(\chi_X=(f^2+g^2)\cdot\frac{1}{f^2+g^2}\in\langle f,g\rangle\). So the ideal \(\langle f,g\rangle\) contains the identity element of the ring and thus \[C(X)=\langle f,g\rangle\subseteq\langle I_x,g\rangle\subseteq C(X).\] Hence, \(I_x\) is a maximal ideal as claimed. We have established that \(\{I_x\}_{x\in X}\subseteq\mathscr{M}\). It turns out that this is method is quite clumsy. A quicker way to establish that \(I_x\) is maximal is to notice that it is the kernel of the evaluation homomorphism \(C(X)\to\mathbb{R}\) that maps \(f\mapsto f(x)\). Since the homomorphism is clearly surjective, the first isomorphism theorem tells us that \(C(X)/I_x\cong\mathbb{R}\), which is a field. This immediately tells us that \(I_x\) is maximal. Hence, Urysohn's lemma is not (yet) required. The fact that \(\{I_x\}_{x\in X}\subseteq\mathscr{M}\) is purely algebraic. We want to establish the reverse inclusion as well. This is tantamount to showing that every maximal ideal of \(C(X)\) is of the form \(I_x\) for an appropriate choice of \(x\in X\). Let us study a "rogue" maximal ideal \(I\) that is not of the form \(I_x\) for any \(x\in X\). Since \(I\) is maximal and \(I_x\) is maximal for every \(x\in X\), the containment \(I\subseteq I_x\) would immediately imply \(I=I_x\). Hence, \(I\) is not contained in any ideal of the form \(I_x\). This means that for each \(x\in X\), there exists \(f_x\in I\) such that \(f_x(x)\neq0\). For each \(x\in X\), by the continuity of each \(f_x\) and the fact that \(f_x(x)\neq0\), there exists an open neighborhood \(U_x\) of \(x\) such that \(0\notin f_x(U_x)\). This gives us an open cover \(\{U_x\}_{x\in X}\) (notice that to form this open cover, we are invoking the axiom of choice). By compactness, we may extract a finite subcover \(\left\{U_{x_j}\right\}_{j=1}^{n}\). By construction, for each \(x\in X\), there exists at least one \(1\leq j\leq n\) such that \(f_{x_j}(x)\neq0\). So the functions \(f_{x_1},\dots,f_{x_n}\) have no common zero. This means that the function \(f_{x_1}^2+\dots+f_{x_n}^2\) is always positive and so \(\frac{1}{f_{x_1}^2+\dots+f_{x_n}^2}\) is a well-defined continuous function on \(X\). Since \(f_{x_1}^2+\dots+f_{x_n}^2\in I\), we have that \(\chi_X=(f_{x_1}^2+\dots+f_{x_n}^2)\cdot\frac{1}{f_{x_1}^2+\dots+f_{x_n}^2}\in I\). This is a contradiction: no maximal ideal is the unit ideal. Hence, no "rogue" maximal ideals exist. This establishes that \(\mathscr{M}=\{I_x\}_{x\in X}\). Notice the paragraph above uses the same sum of squares trick that we used when we clumsily showed that \(\{I_x\}_{x\in X}\subseteq\mathscr{M}\). In particular, we are using the general fact that the ideal generated by any finite collection of functions in \(C(X)\) that share no common zero is the unit ideal. This is what we have essentially proven in the previous paragraph. Now consider the well-defined map \(\varphi\colon X\to\mathscr{M}\) defined by \(\varphi(x)=I_x\). The above establishes that this map is a surjection. A more subtle point is injectivity. This is where we truly need Urysohn's lemma. Pick \(x,y\in X\) to be distinct points. Since compact Hausdorff spaces are normal, and \(\{x\}\) and \(\{y\}\) are disjoint closed sets, by Urysohn's lemma there exists \(f\in C(X)\) such that \(f(x)=0\) and \(f(y)=1\neq0\). This shows that \(I_x\neq I_y\), which establishes that \(\varphi\) is an injection and thus a bijection. We will establish that \(\varphi\) is in fact a homeomorphism. To do this, we will construct a basis for the topology of \(X\) and for the topology of \(\mathscr{M}\), and show that \(\varphi\) induces a bijection between those bases. For each \(f\in C(X)\), define \[U_f=f^{-1}\left(\mathbb{R}\setminus\{0\}\right),\qquad \tilde{U}_f=\left\{I\in\mathscr{M}\colon f\notin I\right\}.\] We claim that \(\{U_f\}_{f\in C(X)}\) and \(\{\tilde{U}_f\}_{f\in C(X)}\) form bases for the topologies on \(X\) and \(\mathscr{M}\), respectively. To check this, we will use the following standard result from point-set topology. A collection of open subsets \(\mathscr{E}\) of a topological space is a basis for the topology if and only if
We continue to establish the claim that \(\{\tilde{U}_f\}_{f\in C(X)}\) forms a basis for the topology on \(\mathscr{M}\). This is easy with a little knowledge of the Zariski topology on the spectrum of a ring. Define \[X_f=\left\{I\in\text{Spec }C(X)\colon f\notin I\right\}.\] It is a standard fact that \(\{X_f\}_{f\in C(X)}\) forms a basis for the Zariski topology. It is also clear that since \(\tilde{U}_f=\mathscr{M}\cap X_f\) for every \(f\in C(X)\), we have that \(\{\tilde{U}_f\}_{f\in C(X)}\) forms a basis for the subspace topology on \(\mathscr{M}\). Finally, we will establish that for every \(f\in C(X)\), we have \(\varphi(U_f)=\tilde{U}_f\). But this can be done in a single line. \[\varphi(U_f)=\left\{I_x\in\mathscr{M}\colon f(x)\neq0\right\}=\left\{I\in\mathscr{M}\colon f\notin I\right\}=\tilde{U}_f.\] We conclude that \(\varphi\) is a homeomorphism. What is interesting is how we employed the assumptions that \(X\) is Hausdorff and compact. Urysohn's lemma was used in a crucial way to establish that \(\varphi\) is injective, and for this we needed that \(X\) is normal (which uses both assumptions). The compactness assumption was used by itself in the proof that \(\varphi\) is surjective (i.e., the proof of the fact that \(\mathscr{M}=\{I_x\}_{x\in X}\)). However, the astute reader may argue that by proving that \(\{U_f\}_{f\in C(X)}\) forms a basis for the topology on \(X\), we accomplished exactly what we wanted to: we found a way to reconstruct the topology of \(X\) given \(C(X)\). In particular, we used the elements of \(C(X)\) to construct a basis for the topology on \(X\). In doing this, we used no assumption on \(X\) at all; we did not use the assumptions that \(X\) is Hausdorff and compact. Indeed, this construction is valid for any topological space. The issue is that the construction relies heavily on an understanding of the individual continuous functions in \(C(X)\). Usually, it is very difficult to compute preimages of arbitrary continuous functions on \(X\). Hence, we would like a better, more direct way to characterize the topology on \(X\). Showing that \(X\) is homeomorphic to \(\mathscr{M}\) (at the expense of some assumptions) gives us a complete picture of the topology (not just a basis) and it relies more on the ring structure of \(C(X)\) than the actual behaviors of the functions in \(C(X)\). From a theoretical point of view, this is a "nicer" characterization of the topology. It is an entirely algebraic characterization. So while it is true that \(C(X)\) always uniquely determines the topology on \(X\), there is an especially nice algebraic way to represent this topology in the case that \(X\) is compact and Hausdorff. This begs the question: what goes wrong with our algebraic characterization when we remove either the assumption of compactness or of being Hausdorff? Since the Hausdorff assumption is a separation axiom, it is fairly intuitive why things may go wrong if it is removed. What is more interesting is if we remove compactness. Let us study what happens when we remove the compactness assumption from a topological subspace \(X\subseteq\mathbb{R}\). By the Heine-Borel theorem, compactness in this context is equivalent to being closed and bounded, so let us separately remove the assumption of being closed and the assumption of being bounded to see what goes wrong in both cases. First, suppose \(X=(0,1)\). This is a set that is bounded but not closed. Let \(J=\left\{f\in C(X)\colon\lim_{y\to 1^-}{f(y)}=0\right\}\). It is easy to check that \(J\) is an ideal. However, it is easy to see that \(J\) is not contained in \(I_x\) for any \(x\in X\) because the function \(g(y)=y-y^2\) is in \(J\) but not in any \(I_x\) since \(g\) is positive on \(X\). Therefore, the maximal ideal containing \(J\) is none of the \(I_x\). So in this case, the inclusion \(\{I_x\}_{x\in X}\subseteq\mathscr{M}\) is strict. Now, suppose that \(X=[0,\infty)\). This is a set that is closed but not bounded. In this case, let \(J=\left\{f\in C(X)\colon\lim_{y\to\infty}{f(y)}=0\right\}\). Once again, this is an ideal. Moreover, the function \(g(y)=e^{-y}\) is in \(J\) but none of the \(I_x\), since \(g\) is positive on \(X\). So in this case as well, the inclusion \(\{I_x\}_{x\in X}\subseteq\mathscr{M}\) is strict. There is one last interesting note. Recall that when we formed an open cover in the argument, we remarked that we were invoking the axiom of choice. This was used to establish that \(\mathscr{M}=\{I_x\}_{x\in X}\). It turns out that that equality can be proven without the axiom of choice using only the assumptions that \(X\) is a complete, totally bounded metric space. See here.
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