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Let $X$ be an uncountable set and let $\mathcal T = \{U \subseteq X : U = \varnothing\text{ or }U^c \text{ is finite} \}$.

Then is topological space $(X,\mathcal T)$

  1. separable?

  2. Hausdorff?

  3. second-countable (has a countable basis)?

  4. first-countable (has a countable basis at each point)?

I am confirmed about (2), $(X,\mathcal T)$ is not a Hausdorff space because we have a result $(X,\mathcal T)$ is Hausdorff iff $D = \{(x,x) : x \in X \}$ is closed , but $D$ is closed if $D^c$ is open and $D^c$ is open if $D^c$ is finite or $\varnothing$ which is not possible . so $(X,\mathcal T)$ is not a Hausdorff.

Please tell me about other three option. Thank you

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Struggler
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1 Answers1

7

Hints:

  1. Show that every infinite set is dense; in particular, the countably infinite sets. (Fix an infinite $A \subseteq X$, and show that $U \cap A \neq \varnothing$ for every nonempty open $U \subseteq X$.)
  2. Show that any two nonempty open sets have nonempty intersection.
  3. If $\mathcal{B} \subseteq \mathcal{T}$ is countable, then the set $A = \bigcup_{U \in \mathcal{B}} ( X \setminus U )$ is countable (since it is a countable union of finite sets). Pick $x \in X \setminus A$, and consider $V = X \setminus \{ x \}$. (Is it a union of sets in $\mathcal{B}$?)
  4. Very similar to the above.

For your attempt at the non-Hausdorffness, you need to be a little bit lot more careful. You need to show that $D$ is not closed in the square $X \times X$. In order to proceed as you have done, you would first have to show that the product $X \times X$ also has the co-fintie topology (since you make an appeal to this). However this is not true. If $A, B \subseteq X$ are finite nonempty, then $( X \setminus A ) \times ( X \setminus B )$ is open in $X \times X$, but it is not co-finite (since for any $a \in A$ the uncountable set $\{ \langle a , x \rangle : x \in X \}$ is disjoint from this set).

user642796
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