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The answer is No.

Let $\eta$ be the order type of $\mathbb{Q}$, and $\omega_1$ - the order type of the set of countable ordinals. The order types $\eta$ and $\eta \cdot \omega_1$ are different (because they have different cardinality), but the set of order types of all proper initial segments of some instances of $\eta$ and $\eta \cdot \omega_1$ is the same. Actually, as proved by Joel David HamkinsJoel David Hamkins, there are $2^{\aleph_1}$ distinct order types with this property: https://math.stackexchange.com/a/174404/19661

The answer is No.

Let $\eta$ be the order type of $\mathbb{Q}$, and $\omega_1$ - the order type of the set of countable ordinals. The order types $\eta$ and $\eta \cdot \omega_1$ are different (because they have different cardinality), but the set of order types of all proper initial segments of some instances of $\eta$ and $\eta \cdot \omega_1$ is the same. Actually, as proved by Joel David Hamkins, there are $2^{\aleph_1}$ distinct order types with this property: https://math.stackexchange.com/a/174404/19661

The answer is No.

Let $\eta$ be the order type of $\mathbb{Q}$, and $\omega_1$ - the order type of the set of countable ordinals. The order types $\eta$ and $\eta \cdot \omega_1$ are different (because they have different cardinality), but the set of order types of all proper initial segments of some instances of $\eta$ and $\eta \cdot \omega_1$ is the same. Actually, as proved by Joel David Hamkins, there are $2^{\aleph_1}$ distinct order types with this property: https://math.stackexchange.com/a/174404/19661

replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Source Link

The answer is No.

Let $\eta$ be the order type of $\mathbb{Q}$, and $\omega_1$ - the order type of the set of countable ordinals. The order types $\eta$ and $\eta \cdot \omega_1$ are different (because they have different cardinality), but the set of order types of all proper initial segments of some instances of $\eta$ and $\eta \cdot \omega_1$ is the same. Actually, as proved by Joel David Hamkins, there are $2^{\aleph_1}$ distinct order types with this property: http://math.stackexchange.com/a/174404/19661https://math.stackexchange.com/a/174404/19661

The answer is No.

Let $\eta$ be the order type of $\mathbb{Q}$, and $\omega_1$ - the order type of the set of countable ordinals. The order types $\eta$ and $\eta \cdot \omega_1$ are different (because they have different cardinality), but the set of order types of all proper initial segments of some instances of $\eta$ and $\eta \cdot \omega_1$ is the same. Actually, as proved by Joel David Hamkins, there are $2^{\aleph_1}$ distinct order types with this property: http://math.stackexchange.com/a/174404/19661

The answer is No.

Let $\eta$ be the order type of $\mathbb{Q}$, and $\omega_1$ - the order type of the set of countable ordinals. The order types $\eta$ and $\eta \cdot \omega_1$ are different (because they have different cardinality), but the set of order types of all proper initial segments of some instances of $\eta$ and $\eta \cdot \omega_1$ is the same. Actually, as proved by Joel David Hamkins, there are $2^{\aleph_1}$ distinct order types with this property: https://math.stackexchange.com/a/174404/19661

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The answer is No.

Let $\eta$ be the order type of $\mathbb{Q}$, and $\omega_1$ - the order type of the set of countable ordinals. The order types $\eta$ and $\eta \cdot \omega_1$ are different (because they have different cardinality), but the set of order types of all proper initial segments of some instances of $\eta$ and $\eta \cdot \omega_1$ is the same. Actually, as proved by Joel David Hamkins, there are $2^{\aleph_1}$ distinct order types with this property: http://math.stackexchange.com/a/174404/19661

The answer is No.

Let $\eta$ be the order type of $\mathbb{Q}$, and $\omega_1$ - the order type of the countable ordinals. The order types $\eta$ and $\eta \cdot \omega_1$ are different (because they have different cardinality), but the set of order types of all proper initial segments of some instances of $\eta$ and $\eta \cdot \omega_1$ is the same. Actually, there are $2^{\aleph_1}$ distinct order types with this property: http://math.stackexchange.com/a/174404/19661

The answer is No.

Let $\eta$ be the order type of $\mathbb{Q}$, and $\omega_1$ - the order type of the set of countable ordinals. The order types $\eta$ and $\eta \cdot \omega_1$ are different (because they have different cardinality), but the set of order types of all proper initial segments of some instances of $\eta$ and $\eta \cdot \omega_1$ is the same. Actually, as proved by Joel David Hamkins, there are $2^{\aleph_1}$ distinct order types with this property: http://math.stackexchange.com/a/174404/19661

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