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If $P$ is a topological property, then a space $(X,\tau)$ is said to be minimal $P$ (respectively, maximal $P$) if $(X,\tau)$ has property $P$ but no topology on $X$ which is strictly smaller (respectively, strictly larger) than $\tau$ has $P$.

We know that if $P$ is topological property, then $P$- minimal will be topological property.

Is $P$ - maximal topological property, if $P$ is topological property?

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I will verfiy that the property $P-$maximal is preserved by homeomorphisms. Let $(X,\tau),(Y,\sigma)$ be topological spaces and $h:X\rightarrow Y$ be a homeomorphism. Let $X$ have a $P-$maximal property for some topological property $P$. Since $X$ has the topological property $P$, therefore $Y$ has the property $P$ as well (because $X\cong Y$). Now suppose that there is topology $\sigma'$ on $Y$ such that $(Y,\sigma')$ has the property $P$ and contains $\sigma$ properly. Verify that $\tau'$ given by $\{h^{-1}[V]|V\in\sigma'\}$ is a topology on $X$ (the bijectivity of $h$ will be used). Verify that $(X,\tau')\cong (Y,\sigma')$. Since $(Y,\sigma')$ has the property $P$, therefore $(X,\sigma')$ has the property $P$. This would mean that $\tau'\subseteq \tau$. Finally verify that this implies that $\sigma'\subseteq \sigma$(bijectivity of $h$ will be needed again here). (contradiction)

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