# Finer topologies on a compact Hausdorff space

If we have such topological space $(X,\mathcal{T})$ that it is compact and Hausdorff, then we can say that for any other topology $\mathcal{H}$ on $X$ such that $\mathcal{T}\subseteq\mathcal{H}$, the topology $\mathcal{H}$ is Hausdorff but no compact.

Hint: Let $(X,\mathcal{T})$, $(Y,\mathcal{H})$ be such two topological spaces that $X$ is compact and $Y$ is Hausdorff. If $f:X\to Y$ is a continuous bijective map, then $f$ is a homeomorphism.

I have tried the following but I don't understand the problem:

Suppose that there exists such topological space $\mathcal{H}$ that $\mathcal{T}\subseteq\mathcal{H}$ and $(X,\mathcal{H})$ is compact, then there exist such countable set $C$ that $\overline{C}=X$.

• Here is an example: Consider $X = [0, 1]$ with $\mathcal T$ being the standard topology and $\mathcal H$ the discrete topology. Is there a mistake in your question? – Ayman Hourieh Jun 9 '14 at 19:55
• Consider the identity map from $(X,\mathcal H)$ to $(X,\mathcal T)$. – David Mitra Jun 9 '14 at 20:00
• @DavidMitra I may be misreading the question, but your hint can be used to show that $\mathcal T$ is compact, not the other way around. – Ayman Hourieh Jun 9 '14 at 20:04
• @AymanHourieh I think the OP wants to show no topology on $X$ that is strictly finer than $\mathcal T$ can be compact. This is addressed here. Is my hint off (if $(X,\mathcal H)$ were compact, the identity would be a homeomorphism, and the two topologies would be the same)? – David Mitra Jun 9 '14 at 20:07
• @DavidMitra But the question says $\mathcal T \subseteq \mathcal H$, i.e. equality is not ruled out. SHB could you clarify your question? – Ayman Hourieh Jun 9 '14 at 20:20

Let $(X, \mathcal {T})$ a compact Hausdorff space and $\mathcal{H}$ a strictly finer topology $\mathcal{T}$, if we take the identity function $i$ such that $i :(X,\mathcal {H}) \to (X, \mathcal {T})$, this function is continuous because $\mathcal{T} \varsubsetneq \mathcal {H}$.     Since $\mathcal {H}$ is strictly finer than $\mathcal {T}$ can take $C$ closed in $\mathcal{H}$ is not in $\mathcal {T}$. If $(X, \mathcal {H})$ was compact would have a compact in $\mathcal {H}$, now as $C=i(C)$ is $C$ would have a compact $\mathcal{T}$ then $\mathcal {T} = \mathcal {H}$ which can not be.