[proofplan] We work inside $L$ and prove the upper bound $|\mathcal{P}(\kappa)\cap L|^L \leq \kappa^+$ for an arbitrary infinite $L$-cardinal $\kappa$. Given $X \subseteq \kappa$ with $X \in L$, we place $X$ in a sufficiently large level $L_\theta$, take an elementary substructure of size $\kappa$ containing $X$ and all ordinals below $\kappa+1$, and collapse it. The Condensation Lemma identifies the collapse with some $L_\beta$ where $\beta < \kappa^+$, so every constructible subset of $\kappa$ already appears before level $L_{\kappa^+}$. Since $L_{\kappa^+}$ has cardinality $\kappa^+$ in $L$, this gives the upper bound, while [Cantor's theorem](/theorems/621) gives the reverse inequality. [/proofplan]

[step:Work in $L$ and fix the cardinal whose power set is being counted] Work in the universe $L$. Let $\kappa$ be an infinite cardinal of $L$, and let $\kappa^+$ denote the successor cardinal of $\kappa$ computed in $L$. Define \begin{align*} \mathcal{P}_L(\kappa) := \{X \in L : X \subseteq \kappa\}. \end{align*} It is enough to prove that $|\mathcal{P}_L(\kappa)|^L = \kappa^+$. Since $L$ satisfies the axioms of choice, cardinal comparison in $L$ is linear, so it suffices to prove both inequalities \begin{align*} \kappa^+ \leq |\mathcal{P}_L(\kappa)|^L \qquad\text{and}\qquad |\mathcal{P}_L(\kappa)|^L \leq \kappa^+. \end{align*} [/step]

[step:Use Cantor's theorem to obtain the lower bound] By Cantor's theorem in $L$ (citing a result not yet in the wiki: Cantor's theorem), there is no surjection in $L$ from $\kappa$ onto $\mathcal{P}_L(\kappa)$. Hence \begin{align*} \kappa < |\mathcal{P}_L(\kappa)|^L. \end{align*} Because $\kappa^+$ is the least $L$-cardinal strictly larger than $\kappa$, it follows that \begin{align*} \kappa^+ \leq |\mathcal{P}_L(\kappa)|^L. \end{align*} [/step]

[step:Show every constructible subset of $\kappa$ appears before $L_{\kappa^+}$]We prove that \begin{align*} \mathcal{P}_L(\kappa) \subseteq L_{\kappa^+}. \end{align*} Let $X \in \mathcal{P}_L(\kappa)$. Since $X \in L$, there is an ordinal $\theta$ such that $X \in L_\theta$. Increase $\theta$ if necessary so that $\theta > \kappa^+$ and $L_\theta$ satisfies a sufficiently large finite fragment of ZFC for the Downward Löwenheim-Skolem argument and the Condensation Lemma. Apply the [Downward Löwenheim-Skolem theorem](/theorems/4272) inside $L$ (citing a result not yet in the wiki: [Downward Löwenheim-Skolem theorem](/theorems/4299)) to the structure $(L_\theta,\in)$ with the parameter set \begin{align*} A := \{X\} \cup (\kappa+1). \end{align*} Since $\kappa$ is infinite, $\kappa+1$ has $L$-cardinality $\kappa$, and adjoining the single element $X$ does not increase that cardinality. Hence $A$ has $L$-cardinality $\kappa$. Thus there is an elementary substructure \begin{align*} M \prec L_\theta \end{align*} such that $A \subseteq M$ and $|M|^L = \kappa$. Let \begin{align*} \pi : M \to N \end{align*} be the Mostowski collapse map, where $N$ is the transitive collapse of $M$. Since $M \prec L_\theta$ and $L_\theta$ is a level of the constructible hierarchy, the Condensation Lemma for $L$ (citing a result not yet in the wiki: Condensation Lemma for constructible levels) gives an ordinal $\beta$ such that \begin{align*} N = L_\beta. \end{align*} Because $|M|^L=\kappa$, also $|L_\beta|^L=\kappa$. In $L$, every ordinal $\alpha \leq \beta$ belongs to $L_\beta$, so the ordinal $\beta$ injects into $L_\beta$. Therefore $|\beta|^L \leq \kappa$, and by the definition of $\kappa^+$ we have \begin{align*} \beta < \kappa^+. \end{align*} Since $\kappa+1 \subseteq M$, the collapse map fixes every ordinal $\alpha \leq \kappa$. Because $X \subseteq \kappa$ and $X \in M$, it follows that \begin{align*} \pi(X) = X. \end{align*} Hence $X \in N = L_\beta \subseteq L_{\kappa^+}$. Since $X \in \mathcal{P}_L(\kappa)$ was arbitrary, \begin{align*} \mathcal{P}_L(\kappa) \subseteq L_{\kappa^+}. \end{align*}[/step]

[guided]The goal is to prove that no subset of $\kappa$ first appears at or above stage $\kappa^+$ of the constructible hierarchy. Fix a constructible subset $X \subseteq \kappa$. Since $X \in L$, there is some ordinal $\theta$ with $X \in L_\theta$. We choose $\theta$ larger than $\kappa^+$ and large enough that $L_\theta$ has the finite amount of set-theoretic structure needed for elementary submodels and condensation. Now define the parameter set \begin{align*} A := \{X\} \cup (\kappa+1). \end{align*} We include $X$ itself as an element of $A$, not merely the members of $X$, because later we must apply the collapse map to $X$. This set has $L$-cardinality $\kappa$: the ordinal $\kappa+1$ has cardinality $\kappa$ because $\kappa$ is infinite, and adjoining the single element $X$ does not increase that cardinality. By the Downward Löwenheim-Skolem theorem inside $L$ (citing a result not yet in the wiki: Downward Löwenheim-Skolem theorem), applied to the structure $(L_\theta,\in)$ and the parameter set $A$, there is an elementary substructure \begin{align*} M \prec L_\theta \end{align*} such that $A \subseteq M$ and $|M|^L=\kappa$. Because $A \subseteq M$, we have both $X \in M$ and $\kappa+1 \subseteq M$. The reason for putting $\kappa+1$ into $M$ is that the collapse must fix every ordinal below or equal to $\kappa$. Let \begin{align*} \pi : M \to N \end{align*} be the Mostowski collapse map. The structure $N$ is transitive, and $\pi$ is an isomorphism from $(M,\in)$ onto $(N,\in)$. Since $M$ is elementary in a constructible level $L_\theta$, the Condensation Lemma for $L$ (citing a result not yet in the wiki: Condensation Lemma for constructible levels) gives an ordinal $\beta$ with \begin{align*} N = L_\beta. \end{align*} We next check that this $\beta$ lies below $\kappa^+$. Since $M$ has $L$-cardinality $\kappa$ and $\pi$ is a bijection from $M$ onto $L_\beta$, we have $|L_\beta|^L=\kappa$. The ordinal $\beta$ injects into $L_\beta$ in $L$, because every ordinal below $\beta$ is an element of $L_\beta$. Therefore $|\beta|^L \leq \kappa$. Since $\kappa^+$ is the least $L$-cardinal larger than $\kappa$, this implies \begin{align*} \beta < \kappa^+. \end{align*} Finally, because $\kappa+1 \subseteq M$, the Mostowski collapse fixes every ordinal $\alpha \leq \kappa$. Since $X \subseteq \kappa$ and $X \in M$, every member of $X$ is fixed, so the image of $X$ under the collapse is exactly $X$: \begin{align*} \pi(X) = X. \end{align*} But $\pi(X) \in N = L_\beta$, hence $X \in L_\beta$. Since $\beta < \kappa^+$, we get $L_\beta \subseteq L_{\kappa^+}$, and therefore $X \in L_{\kappa^+}$. As $X$ was arbitrary, this proves \begin{align*} \mathcal{P}_L(\kappa) \subseteq L_{\kappa^+}. \end{align*}[/guided]

[step:Count the subsets of $\kappa$ using the size of $L_{\kappa^+}$] We use the standard cardinal estimate for constructible levels (citing a result not yet in the wiki: cardinality of constructible levels): for every infinite ordinal $\delta$, \begin{align*} |L_\delta|^L \leq |\delta|^L. \end{align*} Applying this with $\delta=\kappa^+$ gives \begin{align*} |L_{\kappa^+}|^L \leq \kappa^+. \end{align*} Since $\kappa^+ \subseteq L_{\kappa^+}$, the reverse inequality also holds, so \begin{align*} |L_{\kappa^+}|^L = \kappa^+. \end{align*} From the inclusion proved above, \begin{align*} \mathcal{P}_L(\kappa) \subseteq L_{\kappa^+}, \end{align*} we obtain \begin{align*} |\mathcal{P}_L(\kappa)|^L \leq |L_{\kappa^+}|^L = \kappa^+. \end{align*} [/step]

[step:Conclude the Generalised Continuum Hypothesis in $L$] Combining the lower and upper bounds gives \begin{align*} \kappa^+ \leq |\mathcal{P}_L(\kappa)|^L \leq \kappa^+. \end{align*} Therefore \begin{align*} |\mathcal{P}_L(\kappa)|^L = \kappa^+. \end{align*} Since $\mathcal{P}_L(\kappa)$ is exactly the power set of $\kappa$ computed in $L$, this says \begin{align*} (2^\kappa)^L = (\kappa^+)^L. \end{align*} The cardinal $\kappa$ was an arbitrary infinite cardinal of $L$, so $(L,\in)$ satisfies the Generalised Continuum Hypothesis. [/step]