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$\tau_1,\tau_2,\tau_3$ 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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Try \subsetneq for strict inclusions: $\subsetneq$. –  Dylan Moreland Jun 4 '12 at 13:24
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.) –  Willie Wong Jun 4 '12 at 13:25
You probably mean "tau", which is \tau and looks the following: $\tau$. –  Thomas E. Jun 4 '12 at 13:33
thank god! I thought it comes from Prof.Terence Tao. –  La Belle Noiseuse Jun 4 '12 at 13:34
Sure, that special letter $\tau$ named after Terrence Tao. –  GEdgar Jun 4 '12 at 14:14

1 Answer 1

up vote 11 down vote accepted

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

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