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7 changes: 7 additions & 0 deletions spaces/S000017/properties/P000219.md
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---
space: S000017
property: P000219
value: true
---

Let $Y\subseteq X$ with $|Y|=|X|$. Then any bijection $f:Y\to X$ is a homeomorphism.
13 changes: 13 additions & 0 deletions spaces/S000199/properties/P000219.md
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---
space: S000199
property: P000219
value: true
---

Let $Y\subseteq X$ with $|Y|=|X|$.
There is a unique order isomorphism $f:Y\to X$.
Since $X$ is Alexandrov, so is $Y$ with its subspace topology $\tau_Y$.
The smallest $\tau_Y$-neighborhood of a point $y\in Y$ is
$$(\leftarrow,y]_X\cap Y = (\leftarrow,y]_Y,$$
which coincides with the smallest neighborhood of $y$ in the left ray topology of $Y$.
This shows that $f$ is a homeomorphism.
13 changes: 13 additions & 0 deletions spaces/S000200/properties/P000219.md
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---
space: S000200
property: P000219
value: true
---

Let $Y\subseteq X$ with $|Y|=|X|$.
There is a unique order isomorphism $f:Y\to X$.
Since $X$ is Alexandrov, so is $Y$ with its subspace topology $\tau_Y$.
The smallest $\tau_Y$-neighborhood of a point $y\in Y$ is
$$[y,\to)_X\cap Y = [y,\to)_Y,$$
which coincides with the smallest neighborhood of $y$ in the left ray topology of $Y$.
This shows that $f$ is a homeomorphism.
14 changes: 14 additions & 0 deletions spaces/S000217/properties/P000219.md
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---
space: S000217
property: P000219
value: true
---

Let $Y\subseteq X$ with $|Y|=|X|$.
Using well-orderedness and the fact that $\omega_1$ is an initial ordinal,
one can show that there is a unique order isomorphism $f:Y\to X$.
Since $X$ is Alexandrov, so is $Y$ with its subspace topology $\tau_Y$.
The smallest $\tau_Y$-neighborhood of a point $y\in Y$ is
$$(\leftarrow,y]_X\cap Y = (\leftarrow,y]_Y,$$
which coincides with the smallest neighborhood of $y$ in the left ray topology of $Y$.
This shows that $f$ is a homeomorphism.