[proofplan] The proof is a cardinality argument. For each subset $E \subset D$, the sets $E$ and $D \setminus E$ are disjoint and closed in $X$ (since every subset of a closed discrete subspace is closed). By normality, [Urysohn's Lemma](/theorems/887) produces a continuous function $f_E: X \to [0,1]$ separating them. Using the density of a [countable set](/page/Countable%20Set) $Q$, we show that the restriction map $E \mapsto f_E|_Q$ is injective from $\mathcal{P}(D)$ into $[0,1]^Q$. A cardinality comparison then gives $2^{|D|} \le \mathfrak{c}$, from which $|D| \le \mathfrak{c}$ follows. [/proofplan] [step:Show that every subset of the closed discrete subspace $D$ is closed in $X$] Let $D \subset X$ be a closed discrete subspace. Since $D$ carries the discrete topology, every subset $E \subset D$ is closed in $D$. Since $D$ is closed in $X$, the transitivity of closedness gives that $E$ is closed in $X$: if $x \in X \setminus E$, then either $x \in X \setminus D$ (which is open since $D$ is closed) or $x \in D \setminus E$ (which is open in $D$, so $x$ has an open neighbourhood $U$ in $X$ with $U \cap D \subset D \setminus E$, hence $U \cap E = \varnothing$). In both cases, $x$ has an open neighbourhood disjoint from $E$, so $E$ is closed. In particular, for any $E \subset D$, both $E$ and $D \setminus E$ are closed in $X$ and disjoint. [guided] A subspace $D$ of $X$ is **discrete** if every subset of $D$ is open in the subspace topology — equivalently, every singleton $\{d\}$ is open in $D$, which in turn means every subset of $D$ is also closed in $D$. For the proof, we need subsets of $D$ to be closed in $X$, not merely in $D$. This is where the hypothesis that $D$ is *closed* in $X$ is essential. The argument uses the general topological fact: if $D$ is closed in $X$ and $E$ is closed in $D$, then $E$ is closed in $X$. To verify this directly: take $x \in X \setminus E$. There are two cases. **Case 1:** $x \notin D$. Then $X \setminus D$ is an open neighbourhood of $x$ (since $D$ is closed in $X$), and $(X \setminus D) \cap E = \varnothing$ since $E \subset D$. **Case 2:** $x \in D \setminus E$. Since $E$ is closed in $D$, the set $D \setminus E$ is open in $D$. By definition of the subspace topology, there exists an open $U \subset X$ with $U \cap D = D \setminus E$ (or more precisely, $x \in U$ for some open $U \subset X$ with $U \cap D \subset D \setminus E$). Then $U \cap E \subset U \cap D \cap E = \varnothing$. In both cases, $x$ has an open neighbourhood disjoint from $E$, confirming that $E$ is closed in $X$. If $D$ were not closed in $X$, this argument would fail: a subset of $D$ closed in the subspace topology need not be closed in $X$. For instance, the rationals $\mathbb{Q}$ with the discrete subspace topology of $\mathbb{R}$ are not closed in $\mathbb{R}$, and so a subset of $\mathbb{Q}$ that is "closed in $\mathbb{Q}$" need not be closed in $\mathbb{R}$. [/guided] [/step] [step:Apply Urysohn's Lemma to separate $E$ from $D \setminus E$ for each subset $E \subset D$] For each $E \subset D$, the sets $E$ and $D \setminus E$ are disjoint [closed subsets](/page/Closed%20Set) of $X$. Since $X$ is normal, the hypotheses of [Urysohn's Lemma](/theorems/887) are satisfied: there exists a continuous function \begin{align*} f_E: X &\to [0,1] \end{align*} with $f_E|_E \equiv 0$ and $f_E|_{D \setminus E} \equiv 1$. This produces a family $\{f_E\}_{E \in \mathcal{P}(D)}$ indexed by the power set of $D$. [guided] We verify the hypotheses of [Urysohn's Lemma](/theorems/887): 1. **$X$ is normal** — given as a hypothesis of Jones' Lemma. 2. **$E$ and $D \setminus E$ are disjoint closed subsets of $X$** — established in the previous step. The lemma produces a continuous $f_E: X \to [0,1]$ with $f_E \equiv 0$ on $E$ and $f_E \equiv 1$ on $D \setminus E$. A key observation: distinct subsets of $D$ produce genuinely different functions on $D$. If $E_1 \ne E_2$, say $d \in E_1 \setminus E_2$, then $f_{E_1}(d) = 0$ (since $d \in E_1$) while $f_{E_2}(d) = 1$ (since $d \in D \setminus E_2$). So the map $E \mapsto f_E|_D$ is already injective. The subtlety is that we will need injectivity of the *restriction to $Q$*, which requires the [continuity](/page/Continuity) of the functions and the density of $Q$. [/guided] [/step] [step:Restrict to a countable [dense subset](/page/Dense%20Subset) and establish injectivity] Since $X$ is separable, fix a countable dense subset $Q \subset X$. Consider the restriction map \begin{align*} \Phi: \mathcal{P}(D) &\to [0,1]^Q \\ E &\mapsto f_E|_Q. \end{align*} We claim $\Phi$ is injective. Let $E_1, E_2 \subset D$ with $E_1 \ne E_2$. There exists $d \in E_1 \setminus E_2$ (or $d \in E_2 \setminus E_1$; assume the former without loss of generality). Then $f_{E_1}(d) = 0$ and $f_{E_2}(d) = 1$. The continuous function $f_{E_1} - f_{E_2}: X \to \mathbb{R}$ satisfies $(f_{E_1} - f_{E_2})(d) = -1$. By continuity, the preimage $(f_{E_1} - f_{E_2})^{-1}((-\infty, -1/2))$ is a nonempty [open set](/page/Open%20Set) containing $d$. Since $Q$ is dense in $X$, every nonempty open set meets $Q$, so there exists $q \in Q$ with $(f_{E_1} - f_{E_2})(q) < -1/2$. In particular, $f_{E_1}(q) \ne f_{E_2}(q)$, giving $\Phi(E_1) \ne \Phi(E_2)$. [guided] This step uses separability to transfer the pointwise difference on $D$ to a difference visible on $Q$. The argument relies on two properties: 1. **Continuity** of $f_{E_1}$ and $f_{E_2}$: if two continuous real-valued functions differ at a point $d$, they differ on an entire open neighbourhood of $d$ (by continuity of their difference). 2. **Density** of $Q$: every nonempty open set contains a point of $Q$. Together, these ensure that the difference between $f_{E_1}$ and $f_{E_2}$ at $d$ is "witnessed" by some point of $Q$. Specifically, $(f_{E_1} - f_{E_2})(d) = -1$ and $(f_{E_1} - f_{E_2})$ is continuous, so the set $U := \{x \in X : (f_{E_1} - f_{E_2})(x) < -1/2\}$ is open and nonempty (it contains $d$). By density of $Q$, there exists $q \in Q \cap U$, giving $f_{E_1}(q) - f_{E_2}(q) < -1/2$, hence $f_{E_1}(q) \ne f_{E_2}(q)$. One could alternatively use the fact that a continuous function into a Hausdorff space is determined by its values on a dense set: if $f_{E_1}|_Q = f_{E_2}|_Q$, then $f_{E_1} = f_{E_2}$ on all of $X$ (since $Q$ is dense and $[0,1]$ is Hausdorff), which would force $E_1 = E_2$ (since $f_E$ determines $E$ via $E = f_E^{-1}(\{0\}) \cap D$). Either approach gives injectivity. [/guided] [/step] [step:Complete the cardinality bound $|D| \le \mathfrak{c}$] The injection $\Phi: \mathcal{P}(D) \hookrightarrow [0,1]^Q$ gives \begin{align*} 2^{|D|} = |\mathcal{P}(D)| \le |[0,1]^Q|. \end{align*} Since $|Q| \le \aleph_0$ and $|[0,1]| = \mathfrak{c} = 2^{\aleph_0}$, cardinal arithmetic yields \begin{align*} |[0,1]^Q| \le (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \cdot \aleph_0} = 2^{\aleph_0} = \mathfrak{c}. \end{align*} Therefore $2^{|D|} \le \mathfrak{c} = 2^{\aleph_0}$. To conclude $|D| \le \mathfrak{c}$: suppose for contradiction that $|D| > \mathfrak{c}$. By Cantor's theorem, $2^{|D|} > |D| > \mathfrak{c}$, contradicting $2^{|D|} \le \mathfrak{c}$. Therefore $|D| \le \mathfrak{c}$. [guided] The cardinal arithmetic uses the standard identity $\aleph_0 \cdot \aleph_0 = \aleph_0$ (the product of two countable infinite cardinals is countable infinite, provable by the diagonal enumeration of $\mathbb{N} \times \mathbb{N}$). Hence \begin{align*} |[0,1]^Q| = |[0,1]|^{|Q|} \le \mathfrak{c}^{\aleph_0} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \cdot \aleph_0} = 2^{\aleph_0} = \mathfrak{c}. \end{align*} For the final deduction: we have established $2^{|D|} \le \mathfrak{c}$. We want $|D| \le \mathfrak{c}$. Note that one cannot simply "take logarithms": the implication $2^{|D|} \le 2^{\aleph_0} \Rightarrow |D| \le \aleph_0$ would require injectivity of $\kappa \mapsto 2^\kappa$, which is not provable in ZFC (it is consistent with ZFC that $2^{\aleph_0} = 2^{\aleph_1}$). Instead, we use Cantor's theorem ($2^\kappa > \kappa$ for every cardinal $\kappa$) via contrapositive. If $|D| > \mathfrak{c}$, then $|D| \ge \mathfrak{c}^+$ (where $\mathfrak{c}^+$ is the cardinal successor of $\mathfrak{c}$), and by Cantor's theorem: \begin{align*} 2^{|D|} > |D| \ge \mathfrak{c}^+ > \mathfrak{c}, \end{align*} contradicting $2^{|D|} \le \mathfrak{c}$. Therefore $|D| \le \mathfrak{c} = 2^{\aleph_0}$. [/guided] [/step]