not a Hausdorff space although limits of sequences in it are unique

Question

Prove that the co-countable topology on an uncountable set does not make it a Hausdorff space although limits of sequences in it are unique .

---

how can I do that.I have no idea.thanks for your time.

Answer 1

Hint: The only convergent sequences are the eventually constant sequences.

Answer 2

Let $X$ be an uncountable space. In the co-countable topology, since closed sets are countable, two proper closed sets cannot cover the whole space $X$, so two non-empty open sets cannot be disjoint and therefore two different points cannot be separated by open sets.

On the other hand, if you have a sequence $(x_n)_{n∈ℕ}$ in $X$ which converges to at least a limit $x ∈ X$. Then $U = X\setminus\{x_n;\; n ∈ ℕ, x_n ≠ x\}$ is an open neighbourhood of $x$ and there is an $N ∈ ℕ$ such that $∀n ∈ ℕ,\, n > N:\; x_n ∈ U$, meaning $x_n = x$ for all $n > N$. Now for every $y ∈ X$ different from $x$ you have a neighbourhood $V = X\setminus\{x\}$ of $y$ which doesn’t contain $x$. Therefore, the sequence $(x_n)_{n ∈ ℕ}$ can’t converge to $y$, as for all $K ∈ ℕ$ there is an $k ∈ ℕ,\, k > K$, viz $k = \max\{N+1,K+1\}$, such that $x_k = x$ and so $x_k \notin V$.

Answer 3

If you have two open sets in the co-countable topology $U$ and $V$ such that: $$U\cap V= \emptyset$$ Then: $$(U\cap V)^c=\emptyset^c\Longrightarrow U^c\cup V^c=X.$$ Given that $X$ is uncountable, what is wrong with this statement? What is the definition of an open set in the co-countable topology? What is the definition of Hausdorff?