Natural map from topological space to max-spectrum of ring of continuous real-valued functions is an injection iff the space is Urysohn

Statement
Suppose $$X$$ is a topological space. Let $$C(X,\R)$$ be the ring of continuous real-valued functions on $$X$$. Then, consider the following map from $$X$$ to the max-spectrum of $$C(X,\R)$$:

$$x \mapsto M_x = \{ f \in C(X,\R) \mid f(x) = 0 \}$$.

This map is injective if and only if $$X$$ is a fact about::Urysohn space.

Facts used

 * 1) uses::Topological space maps naturally to max-spectrum of ring of continuous real-valued functions

Proof
By fact (1), the map does indeed go to the max-spectrum of $$C(X,\R)$$. So, we show here only that the map is injective if and only if $$X$$ is Urysohn.

Urysohn implies the map is injective
Given: An Urysohn space $$X$$.

To prove: For $$x,y \in X$$, $$M_x \ne M_y$$.

Proof: Since $$X$$ is Urysohn, there exists a continuous function $$f:X \to [0,1]$$ such that $$f(x) = 0$$ and $$f(y) = 1$$. Since $$[0,1]$$ is a subspace of $$\R$$, we can think of $$f$$ as an element in $$C(X,\R)$$. Thus, $$f \in M_x$$ but $$f \notin M_y$$, so we get $$M_x \ne M_y$$.

The map is injective implies Urysohn
Given: A space $$X$$ such that for any two points $$x \ne y \in X$$, $$M_x \ne M_y$$.

To prove: For $$x \ne y \in X$$, there exists a continuous function $$f:X \to [0,1]$$ such that $$f(x) = 0$$ and $$f(y) = 1$$.

Proof: Since $$M_x \ne M_y$$ and both are maximal ideals, neither is contained in the other, so there exists $$g \in M_x$$ such that $$g \notin M_y$$. Thus, $$g(x) = 0$$ and $$g(y) = c \ne 0$$. First, note that we can define a new continuous function $$h \in C(X,\R)$$ such that $$h(p) = g(p)/c$$, so $$h(x) = 0$$ and $$h(y) = 1$$.

Finally consider the function $$f \in C(X,\R)$$ given by $$f(p) = \min \{ |h(p)|, 1 \}$$. Note that $$f$$ is continuous, and its image is in $$[0,1]$$, so we can treat $$f$$ as a function from X to $$[0,1]$$. Thus, we have $$f:X \to [0,1]$$ with $$f(x)=0$$, $$f(y) = 1$$.