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This problem is very much open.

Cheng Yong calls Harrington's $\star$ the assumption that there is a real $x$ such that all $x$-admissible ordinals are $L$-cardinals. From the work of Yong we know that Second- and even Third-order arithmetic do not suffice to prove that Harrington's $\star$ implies the existence of $0^\sharp$. Whether this was possible was originally asked by Woodin, who proved that Second order arithmetic suffices to show that $\mathrm{Det}(\Sigma^1_1)$ implies Harrington's $\star$.

This appears in Yong's dissertation, Analysis of Martin-Harrington theorem "$\mathrm{Det}(\Sigma^1_1)\leftrightarrow 0^\sharp$ exists" in higher order arithmetic, National University of Singapore, 2012.

Here,

• Second order arithmetic, $Z_2$, is formalized as: $(\mathsf{ZFC}-$ Power set axiom$)+$ All sets are countable.
• Third order arithmetic, $Z_3$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P(\omega)$ exists $+$ All sets have size at most continuum.
• Fourth order arithmetic, which Yong shows proves that Harrington's $\star$ implies the existence of $0^\sharp$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P^2(\omega)$ exists $+$ All sets have size at most $2^{2^{\aleph_0}}$.

In joint work by Ralf Schindler and Yong, the situation has been further clarified: $Z_2+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}$, and $Z_3+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}+$ There is a remarkable cardinal. See here.

As question 8.0.14 in his thesis, Yong asks whether $Z_2$ proves Harrington theorem. This is tricky, since $Z_2$ proves that $\mathrm{Det}(\mathbf\Sigma^1_1)$ is equivalent with "For all reals $x$, $x^\sharp$ exists." The reason why the lightface version is open, Yong admits, is that the only route we have from $\mathrm{Det}(\Sigma^1_1)$ to the existence of $0^\sharp$ is via Harrington's $\star$, so the question is effectively asking for a different argument. There are several sufficiently different proofs of Harrington's theorem. For example, there is the argument in Recursive Aspects of Descriptive Set Theory by Mansfield and Weitkamp, via non-$\omega$-models of $\mathsf{KP}$. There is Harrington's original proof, using Steel's forcing. There is Sami's proof, from Analytic determinacy and $0^\sharp$: A forcing-free proof of Harrington's theorem. All use Harrington's $\star$ in an essential fashion.

The question of whether a different route is possible has been studied, of course, but unsuccessfully.

This problem is very much open.

Cheng Yong calls Harrington's $\star$ the assumption that there is a real $x$ such that all $x$-admissible ordinals are $L$-cardinals. From the work of Yong we know that Second- and even Third-order arithmetic do not suffice to prove that Harrington's $\star$ implies the existence of $0^\sharp$. Whether this was possible was originally asked by Woodin, who proved that Second order arithmetic suffices to show that $\mathrm{Det}(\Sigma^1_1)$ implies Harrington's $\star$.

This appears in Yong's dissertation, Analysis of Martin-Harrington theorem "$\mathrm{Det}(\Sigma^1_1)\leftrightarrow 0^\sharp$ exists" in higher order arithmetic, National University of Singapore, 2012.

Here,

• Second order arithmetic, $Z_2$, is formalized as: $(\mathsf{ZFC}-$ Power set axiom$)+$ All sets are countable.
• Third order arithmetic, $Z_3$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P(\omega)$ exists $+$ All sets have size at most continuum.
• Fourth order arithmetic, which Yong shows proves that Harrington's $\star$ implies the existence of $0^\sharp$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P^2(\omega)$ exists $+$ All sets have size at most $2^{2^{\aleph_0}}$.

In Ralf Schindler and Yong's joint work "Harrington's Principle in Higher Order Arithmetic" (published in J. Symb. Log. vol. 80, no. 2), the situation has been further clarified: $Z_2+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}$, and $Z_3+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}+$ There is a remarkable cardinal.

As question 8.0.14 in his thesis, Yong asks whether $Z_2$ proves Harrington theorem. This is tricky, since $Z_2$ proves that $\mathrm{Det}(\mathbf\Sigma^1_1)$ is equivalent with "For all reals $x$, $x^\sharp$ exists." The reason why the lightface version is open, Yong admits, is that the only route we have from $\mathrm{Det}(\Sigma^1_1)$ to the existence of $0^\sharp$ is via Harrington's $\star$, so the question is effectively asking for a different argument. There are several sufficiently different proofs of Harrington's theorem. For example, there is the argument in Recursive Aspects of Descriptive Set Theory by Mansfield and Weitkamp, via non-$\omega$-models of $\mathsf{KP}$. There is Harrington's original proof, using Steel's forcing. There is Sami's proof, from Analytic determinacy and $0^\sharp$: A forcing-free proof of Harrington's theorem. All use Harrington's $\star$ in an essential fashion.

The question of whether a different route is possible has been studied, of course, but unsuccessfully.

This problem is very much open.

Cheng Yong calls Harrington's $\star$ the assumption that there is a real $x$ such that all $x$-admissible ordinals are $L$-cardinals. From the work of Yong we know that Second- and even Third-order arithmetic do not suffice to prove that Harrington's $\star$ implies the existence of $0^\sharp$. Whether this was possible was originally asked by Woodin, who proved that Second order arithmetic suffices to show that $\mathrm{Det}(\Sigma^1_1)$ implies Harrington's $\star$.

This appears in Yong's dissertation, Analysis of Martin-Harrington theorem "$\mathrm{Det}(\Sigma^1_1)\leftrightarrow 0^\sharp$ exists" in higher order arithmetic, National University of Singapore, 2012.

Here,

• Second order arithmetic, $Z_2$, is formalized as: $(\mathsf{ZFC}-$ Power set axiom$)+$ All sets are countable.
• Third order arithmetic, $Z_3$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P(\omega)$ exists $+$ All sets have size at most continuum.
• Fourth order arithmetic, which Yong shows proves that Harrington's $\star$ implies the existence of $0^\sharp$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P^2(\omega)$ exists $+$ All sets have size at most $2^{2^{\aleph_0}}$.

In joint work by Ralf Schindler and Yong, the situation has been further clarified: $Z_2+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}$, and $Z_3+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}+$ There is a remarkable cardinal. See here.

As question 8.0.14 in his thesis, Yong asks whether $Z_2$ proves Harrington theorem. This is tricky, since $Z_2$ proves that $\mathrm{Det}(\mathbf\Sigma^1_1)$ is equiconsistent with "For all reals $x$, $x^\sharp$ exists." The reason why the lightface version is open, Yong admits, is that the only route we have from $\mathrm{Det}(\Sigma^1_1)$ to the existence of $0^\sharp$ is via Harrington's $\star$, so the question is effectively asking for a different argument. There are several sufficiently different proofs of Harrington's theorem. For example, there is the argument in Recursive Aspects of Descriptive Set Theory by Mansfield and Weitkamp, via non-$\omega$-models of $\mathsf{KP}$. There is Harrington's original proof, using Steel's forcing. There is Sami's proof, from Analytic determinacy and $0^\sharp$: A forcing-free proof of Harrington's theorem. All use Harrington's $\star$ in an essential fashion.

The question of whether a different route is possible has been studied, of course, but unsuccessfully.

This problem is very much open.

Cheng Yong calls Harrington's $\star$ the assumption that there is a real $x$ such that all $x$-admissible ordinals are $L$-cardinals. From the work of Yong we know that Second- and even Third-order arithmetic do not suffice to prove that Harrington's $\star$ implies the existence of $0^\sharp$. Whether this was possible was originally asked by Woodin, who proved that Second order arithmetic suffices to show that $\mathrm{Det}(\Sigma^1_1)$ implies Harrington's $\star$.

This appears in Yong's dissertation, Analysis of Martin-Harrington theorem "$\mathrm{Det}(\Sigma^1_1)\leftrightarrow 0^\sharp$ exists" in higher order arithmetic, National University of Singapore, 2012.

Here,

• Second order arithmetic, $Z_2$, is formalized as: $(\mathsf{ZFC}-$ Power set axiom$)+$ All sets are countable.
• Third order arithmetic, $Z_3$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P(\omega)$ exists $+$ All sets have size at most continuum.
• Fourth order arithmetic, which Yong shows proves that Harrington's $\star$ implies the existence of $0^\sharp$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P^2(\omega)$ exists $+$ All sets have size at most $2^{2^{\aleph_0}}$.

In joint work by Ralf Schindler and Yong, the situation has been further clarified: $Z_2+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}$, and $Z_3+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}+$ There is a remarkable cardinal. See here.

As question 8.0.14 in his thesis, Yong asks whether $Z_2$ proves Harrington theorem. This is tricky, since $Z_2$ proves that $\mathrm{Det}(\mathbf\Sigma^1_1)$ is equivalent with "For all reals $x$, $x^\sharp$ exists." The reason why the lightface version is open, Yong admits, is that the only route we have from $\mathrm{Det}(\Sigma^1_1)$ to the existence of $0^\sharp$ is via Harrington's $\star$, so the question is effectively asking for a different argument. There are several sufficiently different proofs of Harrington's theorem. For example, there is the argument in Recursive Aspects of Descriptive Set Theory by Mansfield and Weitkamp, via non-$\omega$-models of $\mathsf{KP}$. There is Harrington's original proof, using Steel's forcing. There is Sami's proof, from Analytic determinacy and $0^\sharp$: A forcing-free proof of Harrington's theorem. All use Harrington's $\star$ in an essential fashion.

The question of whether a different route is possible has been studied, of course, but unsuccessfully.

This problem is very much open.

Cheng Yong calls Harrington's $\star$ the assumption that there is a real $x$ such that all $x$-admissible ordinals are $L$-cardinals. From the work of Yong we know that Second (or even Third) order arithmetic does not suffice to prove that Harrington's $\star$ implies the existence of $0^\sharp$. Whether this was possible was originally asked by Woodin, who proved that Second order arithmetic suffices to show that $\mathrm{Det}(\Sigma^1_1)$ implies Harrington's $\star$.

This appears in Yong's dissertation, Analysis of Martin-Harrington theorem "$\mathrm{Det}(\Sigma^1_1)\leftrightarrow 0^\sharp$ exists" in higher order arithmetic, National University of Singapore, 2012.

Here,

• Second order arithmetic, $Z_2$, is formalized as: $(\mathsf{ZFC}-$ Power set axiom$)+$ All sets are countable.
• Third order arithmetic is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P(\omega)$ exists $+$ All sets have size at most continuum.
• Fourth order arithmetic, which Yong shows proves that Harrington's $\star$ implies the existence of $0^\sharp$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P^2(\omega)$ exists $+$ All sets have size at most $2^{2^{\aleph_0}}$.

Ralf Schindler has further clarified the situation: $Z_2+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}+$ There is a remarkable cardinal.

As question 8.0.14 in his thesis, Yong asks whether $Z_2$ proves Harrington theorem. This is tricky, since $Z_2$ proves that $\mathrm{Det}(\mathbf\Sigma^1_1)$ is equiconsistent with "For all reals $x$, $x^\sharp$ exists." The reason why the lightface version is open, Yong admits, is that the only route we have from $\mathrm{Det}(\Sigma^1_1)$ to the existence of $0^\sharp$ is via Harrington's $\star$, so the question is effectively asking for a different argument. There are several sufficiently different proofs of Harrington's theorem. For example, there is the argument in Recursive Aspects of Descriptive Set Theory by Mansfield and Weitkamp, via non-$\omega$-models of $\mathsf{KP}$. There is Harrington's original proof, using Steel's forcing. There is Sami's proof, from Analytic determinacy and $0^\sharp$: A forcing-free proof of Harrington's theorem. All use Harrington's $\star$ in an essential fashion. The question of whether a different route is possible has been studied, of course, but unsuccessfully.

This problem is very much open.

Cheng Yong calls Harrington's $\star$ the assumption that there is a real $x$ such that all $x$-admissible ordinals are $L$-cardinals. From the work of Yong we know that Second- and even Third-order arithmetic do not suffice to prove that Harrington's $\star$ implies the existence of $0^\sharp$. Whether this was possible was originally asked by Woodin, who proved that Second order arithmetic suffices to show that $\mathrm{Det}(\Sigma^1_1)$ implies Harrington's $\star$.

This appears in Yong's dissertation, Analysis of Martin-Harrington theorem "$\mathrm{Det}(\Sigma^1_1)\leftrightarrow 0^\sharp$ exists" in higher order arithmetic, National University of Singapore, 2012.

Here,

• Second order arithmetic, $Z_2$, is formalized as: $(\mathsf{ZFC}-$ Power set axiom$)+$ All sets are countable.
• Third order arithmetic, $Z_3$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P(\omega)$ exists $+$ All sets have size at most continuum.
• Fourth order arithmetic, which Yong shows proves that Harrington's $\star$ implies the existence of $0^\sharp$, is $(\mathsf{ZFC}-$ Power set axiom$)+\mathcal P^2(\omega)$ exists $+$ All sets have size at most $2^{2^{\aleph_0}}$.

In joint work by Ralf Schindler and Yong, the situation has been further clarified: $Z_2+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}$, and $Z_3+$ Harrington's $\star$ is equiconsistent with $\mathsf{ZFC}+$ There is a remarkable cardinal. See here.

As question 8.0.14 in his thesis, Yong asks whether $Z_2$ proves Harrington theorem. This is tricky, since $Z_2$ proves that $\mathrm{Det}(\mathbf\Sigma^1_1)$ is equiconsistent with "For all reals $x$, $x^\sharp$ exists." The reason why the lightface version is open, Yong admits, is that the only route we have from $\mathrm{Det}(\Sigma^1_1)$ to the existence of $0^\sharp$ is via Harrington's $\star$, so the question is effectively asking for a different argument. There are several sufficiently different proofs of Harrington's theorem. For example, there is the argument in Recursive Aspects of Descriptive Set Theory by Mansfield and Weitkamp, via non-$\omega$-models of $\mathsf{KP}$. There is Harrington's original proof, using Steel's forcing. There is Sami's proof, from Analytic determinacy and $0^\sharp$: A forcing-free proof of Harrington's theorem. All use Harrington's $\star$ in an essential fashion. 

The question of whether a different route is possible has been studied, of course, but unsuccessfully.