Construction of the Čech-Stone compactification: why is $\iota$ an embedding? I have been reading a post on Terence Tao's blog about the Čech-Stone compactification.  He constructs a compactification as follows (see his exercise 3).  Let $X$ be a locally compact Hausdorff space.  Let $C\left(X,\left[0,1\right]\right)$ be the set of continuous functions from $X$ to $\left[0,1\right]$.  Let $Q=\left[0,1\right]^{C\left(X,\left[0,1\right]\right)}$ with the product topology. Let $\iota\colon X\rightarrow Q$ be given by $\iota\left(x\right)=\left(f\left(x\right)\right)_{f\in C\left(X,\left[0,1\right]\right)}$ for all $x\in X$.  Let $\beta X$ be the closure of $\iota\left(X\right)$ in $Q$.  Then $\left(\beta X,\iota\right)$ is a Čech-Stone compactification of $X$.
I'm trying to see why $\left(\beta X, \iota \right)$ is a compactification.  I am able to show that $\beta X$ is compact and that $\iota$ is a continuous bijection whose image is a dense subset of $\beta X$, but I have been unable to see why $\iota$ must be an embedding. In particular, suppose $X=\mathbb{N}$.  Then $\iota\left(\{0\} \right) = \{\left(f\left(0\right)\right)_{f\in C\left(\mathbb{N}\rightarrow \left[0,1\right]\right)}\}$, which is not, it seems to me, open in $\beta X$ as a subspace of $Q$ with the product topology.  I am sure I am misunderstanding something, as my topology experience is limited.
Edit
My counterexample of $\iota\left(\{0\}\right)$ is wrong, as explained in the comments of @user254665's answer.  Implicitly, I was assuming that $\iota\left(\mathbb{N}\right) = \Pi_{f\in C\left(\mathbb{N} \rightarrow \left[0,1\right]\right)}f\left(\mathbb{N}\right)$, which is incorrect.  Moreover, the following argument shows that $\iota\left(\{0\}\right)$ is in fact open in $\iota\left(\mathbb{N}\right)$.  Let $f \colon \mathbb{N} \rightarrow \left[0,1\right]$ be given by $f\left(0\right)=1$ and $f\left(n\right)=0$ for $n\ne 0$. Let $U = \{q\in Q \vert q_{f}=1\}$.  Then $U$ is open in $\iota\left(\mathbb{N}\right) \subset Q$, $\iota\left(0\right) \in U$, and if $n\ne 0$ then $\iota\left(n\right) \notin U$.

Regardind the Q of why it is an embedding: Apply the Diagonal Theorem.
Let $F=\{f_s:A\to B_s\}_{s\in S}$ be a set of  functions.
We say $F$ separates points when there exists $f\in F$ with $f(a)\ne f(a')$ whenever $a,a'$ are distinct members of $A.$ We say $F$ separates closed sets from points when there exists  $f\in F$ with $f(a)\not \in Cl_{B_s}f(C)$ whenever $a\in A$ and $a\not \in C=Cl_A(C)\subset A.$ 
Theorem. If $F$ (as above) is a set of continuous functions, and separates points, and separates closed sets from points, then $\Delta (a)=(f_s(a))_{s\in S}$ is a homeomorphic embedding of $A$ into $\prod_{s\in S}B_s.$
See, for example, Engelking, General Topology, chapter 2: The paragraphs after Corollary 2.3.18, and Lemma 2.3.19 and Theorem 2.3.20.
