Let $T=\{ U\subseteq X:X\setminus U\textrm{ is countable}\}\cup \{\emptyset\}$

Then this is known as co-countable topology.

Clearly,real line with co-countable topology is not Hausdorff.

For: If T is Hausdorff there exist two disjoint open sets that seperate the any pair of points;say G and H. Then $X\setminus (A \cap B)=X$. So the LHS is countable, but RHS is uncountable, which is a contradiction.

Can I use the same argument to prove the result if X is the complex plane or the Euclidean space?

  • $\begingroup$ You need to add the empty set to $T$, otherwise it's not a topology on $\Bbb R$. This also makes your argument not work even on $\Bbb R$. $\endgroup$
    – nonpop
    Dec 27 '12 at 11:49
  • 1
    $\begingroup$ I've added $\{\emptyset\}$ to your $T$. As nonop mentioned, it is essential, but this also makes it necessary to mention that $A$ and $B$ are both nonempty (which is obvious, but should be mentioned in such a basic exercise anyway). Also, to write tex formulas, encase them in dollar signs ($), like usual. $\endgroup$
    – tomasz
    Dec 27 '12 at 12:05

All that matters for topological properties of the co-countable topology is the cardinality of the underlying set. (It is very easy to show that if $X$ and $Y$ are co-countable topological spaces and $|X| = |Y|$, then $X \cong Y$.)

So, to answer your question: yes, you can use the same argument for these sets, since they all have the same cardinality as $\mathbb{R}$.

But, even more, if $X$ is any uncountable set, then your argument shows that the co-countable topology on $X$ is not Hausdorff (you never used anything particular about the cardinality of $\mathbb{R}$ in your argument, except that it is uncountable).

  • $\begingroup$ @ Arthur,Thank you very much! $\endgroup$
    – ccc
    Dec 27 '12 at 12:19

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