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Let $X$ be an uncountable set with the co-countable topology, and let $A$ be an uncountable subset of $X$. Then $\overline{A}=X$. Show that for each $x \in X \setminus A$ there is no sequence in $A$ that converges to $x$.

Definitions

  • Given an uncountable set $X$, the co-countable topology on $X$ is $$T_{\text{co-countable}} =\{ A \subset X : A^c \cap X \text{ is countable }\} \cup \{ \emptyset \}.$$

  • Let $(X,T)$ be a topological space. If $(x_n)_{n=1}^\infty$ is a sequence in $X$ we say it converges to $x_0 \in X$, $x_n \to x_0$, if for each open set $U \subset X$ containing $x_0$ there is an $n_0 \in \mathbb N$ such that $x_n \in U$ for all $n \geq n_0$.

Using the hint:

Let $x \in X \setminus A \equiv X \cap A^c$. (Can I say $x$ is in a neighborhood of topology space?) So $x \in X $ and $x \in A^c$.

(Guessing will get a contradiciotion somewhere.)

Somehow that it there is a seq $x_n $ that converges to $x$.

I'm not sure where to go from there is there confused.

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    Hint: A sequence is countable, so its complement is an open set. – 2017-02-15

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Proof by contradiction is certainly appropriate here. Picking $x \in X \setminus A$ is a good start, but there's not going to be any contradiction from that - unless $A$ is all of $X$, there will be such an $x$. So then suppose for contradiction that there is a sequence $(x_n)_{n=1}^{\infty}$ converging to $x$.

What does that mean? It means that for every open $N$ containing $x$, there is an $n$ so that $x_n \in N$. Translating that, this means that in every co-countable set $N$ containing $x$, there is an $n$ so that $x_n \in N$.

Can you think of a co-countable set that contains $x$ but not any $x_n$? Equivalently, can you think of a countable set that contains every $x_n$ but not $x$? Such a set would contradict the above. Remember that you can assume $x \neq x_n$, because the $x_n$ are in $A$ and $x$ is not.