[proofplan] We prove the contrapositive estimate: too many prime factors force the integer to be too large. Write the prime factorization of $a$ with multiplicity. Since every prime divisor is at least $X^{1/u}$, the product of $k$ prime factors is at least $X^{k/u}$. If $k \geq r+1$ and $r \geq \alpha u$, then this lower bound exceeds $X^\alpha$, contradicting the assumed size bound $a \leq X^\alpha$. [/proofplan] [step:Fix the sifted integer and write its prime factorization with multiplicity] Fix an element $a \in A$ such that no prime divisor $p$ of $a$ satisfies $p < X^{1/u}$. Fix an integer $r$ with $r \geq \alpha u$. Let $k := \Omega(a)$ denote the number of prime factors of $a$ counted with multiplicity. Since $a \geq 2$, the [Fundamental Theorem of Arithmetic](/theorems/730) gives a prime factorization $a = \prod_{j=1}^{k} p_j$, where each $p_j$ is prime and the primes are listed with multiplicity. By the siftedness hypothesis, each prime $p_j$ satisfies $p_j \geq X^{1/u}$. [guided] We choose one integer $a$ satisfying the hypotheses because the theorem asserts the conclusion separately for each such element of $A$. We also fix an integer $r$ with $r \geq \alpha u$, because we must prove that this same $a$ is a $P_r$ for every such $r$. Define $k := \Omega(a)$. This means that $k$ is the total number of prime factors of $a$, counted with multiplicity. Since $a \geq 2$, the [Fundamental Theorem of Arithmetic](/theorems/730) gives the prime factorization $a = \prod_{j=1}^{k} p_j$, where each $p_j$ is prime and repeated prime factors appear repeatedly in the list. The hypothesis that $a$ has no prime divisor below $X^{1/u}$ applies to every prime factor in this list, so for every index $j \in \{1,\dots,k\}$ we have $p_j \geq X^{1/u}$. [/guided] [/step] [step:Show that more than $r$ prime factors violate the size bound] Suppose, for contradiction, that $a$ is not a $P_r$. By definition of $P_r$, this means $\Omega(a) > r$, so $k \geq r+1$ because $k$ and $r$ are integers. Using the lower bound for each prime factor, we obtain \begin{align*}a = \prod_{j=1}^{k} p_j \geq \prod_{j=1}^{k} X^{1/u} = X^{k/u}.\end{align*} Since $k \geq r+1$ and $X > 1$, we have $X^{k/u} \geq X^{(r+1)/u}$. Since $r \geq \alpha u$ and $u > 0$, we have $(r+1)/u > \alpha$. Because $X > 1$, exponentiation by base $X$ is strictly increasing, hence $X^{(r+1)/u} > X^\alpha$. Combining the inequalities gives $a > X^\alpha$, contradicting the hypothesis $a \leq X^\alpha$. Therefore $\Omega(a) \leq r$, so $a$ is a $P_r$. [guided] Assume toward a contradiction that $a$ is not a $P_r$. The definition of $P_r$ is exactly the inequality $\Omega(a) \leq r$, so failure of this property gives $\Omega(a) > r$. Since $\Omega(a) = k$ and both $k$ and $r$ are integers, this strict inequality is the same as $k \geq r+1$. Now multiply the lower bounds for the prime factors. For every $j \in \{1,\dots,k\}$, we have $p_j \geq X^{1/u}$. Therefore \begin{align*}a = \prod_{j=1}^{k} p_j \geq \prod_{j=1}^{k} X^{1/u} = X^{k/u}.\end{align*} The point of counting prime factors with multiplicity is that the product has exactly $k$ factors, so the lower bound contains the exponent $k/u$. Since $k \geq r+1$ and $X > 1$, increasing the exponent increases the power of $X$, so $X^{k/u} \geq X^{(r+1)/u}$. The condition on $r$ now forces this exponent past the permitted size exponent. Indeed, from $r \geq \alpha u$ and $u > 0$, we get $r+1 > \alpha u$, and division by the positive number $u$ gives $(r+1)/u > \alpha$. Again using $X > 1$, exponentiation with base $X$ is strictly increasing, so $X^{(r+1)/u} > X^\alpha$. Combining these inequalities yields $a > X^\alpha$. This contradicts the assumed size bound $a \leq X^\alpha$ for elements of $A$. Hence the contradiction assumption was false, so $\Omega(a) \leq r$. By definition, $a$ is a $P_r$. [/guided] [/step] [step:Conclude the statement for every admissible $r$] The element $a \in A$ and the integer $r \geq \alpha u$ were arbitrary subject to the stated hypotheses. Therefore every $a \in A$ with no prime divisor below $X^{1/u}$ is a $P_r$ for every integer $r \geq \alpha u$. [/step]