$\tau_1,\tau_2,\tau_3$ are topologies on a set such that $\tau_1\subset \tau_2\subset \tau_3$ and $(X,\tau_2)$ is a compact Hausdorff space. Could any one tell me which of the following are correct?

  1. $\tau_1=\tau_2$ if $(X,\tau_1)$ is compact Hausdorff.
  2. $\tau_1=\tau_2$ if $(X,\tau_1)$ is compact.
  3. $\tau_2=\tau_3$ if $(X,\tau_3)$ is Hausdorff.
  4. $\tau_2=\tau_3$ if $(X,\tau_3)$ is compact.
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    $\begingroup$ Try \subsetneq for strict inclusions: $\subsetneq$. $\endgroup$ – Dylan Moreland Jun 4 '12 at 13:24
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    $\begingroup$ Isn't it slightly problematic to postulate that $\pi_1 \neq \pi_2$ and ask about when can $\pi_1 = \pi_2$? (Now fixed by the community editors [thanks!] but I'm leaving it up for the benefit of the OP.) $\endgroup$ – Willie Wong Jun 4 '12 at 13:25
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    $\begingroup$ You probably mean "tau", which is \tau and looks the following: $\tau$. $\endgroup$ – T. Eskin Jun 4 '12 at 13:33
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    $\begingroup$ thank god! I thought it comes from Prof.Terence Tao. $\endgroup$ – Marso Jun 4 '12 at 13:34
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    $\begingroup$ Sure, that special letter $\tau$ named after Terrence Tao. $\endgroup$ – GEdgar Jun 4 '12 at 14:14

Hint: The identity mapping $(X,\tau_{i+1}) \to (X,\tau_i)$ is continuous and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. This takes care of two statements and the two others are refuted by considering the trivial and the discrete topology on an infinite compact Hausdorff space $(X,\tau_2)$.

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  • $\begingroup$ You meant "continuous 1-to-1 map" here, and 1-to-1ness is clear. $\endgroup$ – hardmath Jun 4 '12 at 15:49
  • $\begingroup$ @hardmath: thanks, fixed. $\endgroup$ – t.b. Jun 4 '12 at 15:49

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