I'm reading Armstrong's Basic Topology and Theorem 3.7 says:

A one-one, onto, and continuous function from a compact space $X$ to a Hausdorff space $Y$ is a homeomorphism.

I was trying to convince myself with examples. Then I thought of this example, which seems a little weird:

Consider the space $X = \{a, b, c\}$. Let $X_1$ be the discrete topology of $X$, and let $X_2$ be the topology $(X, \{\{a\}, \{b\}, \{c\}, \{a, b\}, \{b,c \}, \{a, c\},\{a, b, c\}, \varnothing\})$. Since every pair of distinct points in $X_1$, $X_2$ have disjoint neighborhood, $X_1$ and $X_2$ are Hausdorff. In addition $X_1$ is also compact, since its topology contains finitely many open subsets of $X$. Then define a function $f: X_1 \rightarrow X_2$ such that $f(a) = a, f(b) = b, f(c) = c$.

Obviously, $f$ is onto, one-one, and continuous, but $f^{-1}$ is not continuous(?), since the open set $\{a, b\} \in X_1$ is not mapped to any open set in $X_2$. So $f$ is not a Homeomorphism, which contradicts the theorem (what???).

Assuming this theorem is true and well-phrased, there is definitely something wrong with my reasoning here. So I appreciate any correction of my thought! Thank you a lot!

  • $\begingroup$ Have you taken a look at the proof of the theorem? If you can understand the proof, that may help you clear up confusions. $\endgroup$ – Future Jan 8 '16 at 19:56
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    $\begingroup$ The open set $\{a,b\}$ in $X_1$ is mapped to the open set $\{a,b\}$ in $X_2$ (this set is open in the topology you gave for $X_2$). In fact, the topology you described for $X_2$ is the discrete topology, so any function on $X_2$ is continuous. $\endgroup$ – kccu Jan 8 '16 at 19:57
  • $\begingroup$ @kccu Yes! Oh I forgot that for any topology $T$, the union of $t \in T$ must still be in $T$, thus $X_2$ should in fact have the discrete topology! Thanks for pointing that out! $\endgroup$ – nekodesu Jan 8 '16 at 20:21
  • $\begingroup$ Where you wrote $(X, \{\{a\}, \{b\}, \{c\}, \{a, b\}, \{b,c \}, \{a, c\},\{a, b, c\}, \varnothing\})$, might you have intended $(X, \{a\}, \{b\}, \{c\}, \{a, b\}, \{b,c \}, \{a, c\},\{a, b, c\}, \varnothing )$. That would be a topology on $X$ (provided it is taken to mean the set $\{X, \{a\}, \{b\}, \{c\}, \{a, b\}, \{b,c \}, \{a, c\},\{a, b, c\}, \varnothing\}$ rather than a tuple (as suggested by the use of round brackets), but it would be identical to $X_1$, the discrete topology, so your mapping would be a homeomorphism. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 8 '16 at 20:29

The topology you gave for $X_2$ includes $\{a, b\}$ as an open subset by definition so $f(\{a, b\}) = \{a, b\}$ is absolutely open. If you're trying to see why this is true in general, it will be helpful to recall what we know about compact subspaces of Hausdorff spaces and that to show something is a homeomorphism, it suffices to show that it's a closed map as well.

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  • $\begingroup$ You are absolutely correct -- I forgot that $X_2$ should contain $\{a, b\}$ to have a legit topology. $\endgroup$ – nekodesu Jan 8 '16 at 20:22

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