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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 $\overline{C}=X$.

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  • $\begingroup$ 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? $\endgroup$
    – Ayman Hourieh
    Commented Jun 9, 2014 at 19:55
  • $\begingroup$ Consider the identity map from $(X,\mathcal H)$ to $(X,\mathcal T)$. $\endgroup$
    – David Mitra
    Commented Jun 9, 2014 at 20:00
  • $\begingroup$ @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. $\endgroup$
    – Ayman Hourieh
    Commented Jun 9, 2014 at 20:04
  • $\begingroup$ @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)? $\endgroup$
    – David Mitra
    Commented Jun 9, 2014 at 20:07
  • $\begingroup$ @DavidMitra But the question says $\mathcal T \subseteq \mathcal H$, i.e. equality is not ruled out. SHB could you clarify your question? $\endgroup$
    – Ayman Hourieh
    Commented Jun 9, 2014 at 20:20

1 Answer 1

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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.

am I right?

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