- We will first show a strong nullstellensatz like theorem, showing that every maximal ideal $m$ of the ring of continuous functions $C(x)$ is in bijection with the set of functions that vanish at a point, $V(x)$.

- Let $C(X)$ be the ring of all continuous real valued functions on $X$. For each $x \in X$, define$I(x) \subseteq C(X)$ to be the set of functions that vanish at $x$. This is a maximal ideal, because it is the kernel of the evaluation map $f \mapsto f(x)$.

- Given some maximal ideal $m \subseteq C(X)$, we will show that there is some point $p \in X$ such that$m = I(p)$. To show this, first consider the common zeros of functions in $m$, $V(m) \equiv \{ x \in X : f(x) = 0 \forall f \in m \}$.We first show that $V(m)$ is non-empty, and we then show that $V(m)$ contains exactly one point.

- To show $V(m)$ is non-empty, suppose for contradiction that $V(m)$ is empty. Thus, for each point $x \in X$,not all functions in $m$ vanish at $x$ (otherwise $x \in V(m)$).So, there is a function $f_x \in m$ that does not vanish at $x$, hence $f_x(x) \neq 0$. Since $f_x$is continuous, there is some open neighbourhood $x \in U_x$ where $f(U_x) \neq 0$. (A continuous functionthat does not vanish at a point cannot "suddenly" decay to zero. It will be non-zero over an open nbhd).Since the space $X$ is compact, we have a finite number of $U_{x_i}$ that cover $X$. Hence, we build a function$c \equiv \sum_i f_{x_i}^2$ ($c$ for contradiction) that vanishes nowhere. This means $c$ is a unit.But we must have $c \in m$ as $c$ is built out of functions $f_{x_i} \in m$. This is a contradiction as a unitcannot belong to a maximal ideal. Thus, $V(m)$ contains at least one point.

- To show that $V(m)$ contains exactly one point, suppose that $V(m)$ contains a single point $x$.This means that all functions in $m$ vanish at $x$. Thus, $m \subseteq I(x)$, since $I(x)$ contains all functions(not just ones in $m$) that vanish at $x$. But $m$ is maximal, and hence $m = I(x)$. This tells us thatevery maximal ideal $m$ corresponds to some vanishing set $I(x)$.

- We will next show that every vanishing set $I(x)$ is distinct. We already know that it is maximal. This gives us an injection. Let $I(p), I(q)$ be two vanishing sets for distinct points. Let $z_p$ be the function constructed fromUrhyson's lemma that is zero at $p$ at nonzero at $q$. Thus, we have $z_p \in I(p)$ and $z_p \not \in I(q)$. Hence,$I(p) \neq I(q)$. This shows that the maximal ideals $I(p), I(q)$ will be distinct.

- We have thus established a
/*bijection*between zero sets maximal ideals $V(m)$ and functions that vanish at a point $I(p)$.*nullstellensatz*

- We will next show that this provides a homeomorphism. It suffices to consider basic open sets. We know that thesets $D_{spec}(f) = \{ m \in C(X) : f \not \in m \}$ is a basis for the maximal spectrum of the ring under zariski.We will show that $D_{top}(f) \equiv \{ x \in X: f(x) \neq 0 \}$ is a basis for the topology of $X$. Then the functionthat takes points to maximal ideals of functions that vanish at that point will provide a topological homeomorphism.Thus, we have shown that the maximal spectrum of the ring allows us to recover the topology of the underlying space!

- We wish to show that the open set $D_{top}(f)$ form a base for the topology on $X$. So consider an open set $U \subseteq X$. Now think of $U^c$ which is closed. We build the function $d(x, U)$ such that $d(x, U)(x) = 1$and $d(x, U)(U^c) = 0$ by invoking Urhyson. Therefore, $x \in D_{top}(d(x, U)) \subseteq U$. So the set $U$ can becovered with $\{ D_{top}(d(x, U)): x \in U \}$, which means the sets $D(d(x, U))$ form a base of the topology on $X$.

- We wish to show that the open sets $D_{spec}(f)$ form a base for the topology on $maxSpec(C(X))$. Let $U$ be aclosed set in $maxSpec(C(X))$.

- We wish to show that the open set $D_{spec}(f)$ have homeomorphisms $D_{top}(f)$. This completes the isomorphism into a homeomorphism, and we have thus completed the proof that we can recover the topology from the spectrum.

- Consider the function $zero: X \rightarrow mSpec(C(X))$ sending the point $x$ to the kernel of the evaluation map at $x$.Let $D_{spec}(f) \subseteq mSpec(C(X))$ be a basic open of $mSpec(C(X))$. Consider $zero^{-1}(D_{spec}(f))$.This will contain all those points $x \in X$ such that $zero(x) \in D_{spec}(f)$. This means that it will containpoint $x \in X$ such that $f$ does not vanish at those points, as (1) $zero(x) \in D_{spec}(f)$ implies(2) $f \not \in zero(x)$ which implies $f(x) \neq 0$. Clearly, this is an open subset of $X$, as it is thecomplement of the closed set $f(x) = 0$ [zero sets are always closed]. Furthermore, the set $zero^{-1}(D_{spec}(f))$maps to what we would expect; it trades the algebraic definition of "does not vanish" to the geometric one,while describing the exact same phenomena.