Construct $X$ s.t. it has a dense subset but elements in the complement of the subset have no sequence in the subset converging to them

Question

Construct a topological space $X$ with the following property: There is a proper subset $Y$ of $X$ such that the closure of $Y$ is $X$ but if $c \in X-Y$ then there is no sequence in $Y$ converging to $c$.

I believe this would mean that the space I'm looking for can not be first countable, because first countable is what gaurentee's elements in the closure of a subset to have sequences in subset converging to them. That being said, I'm at a bit of a loss. I guess my first thought is it can't be a metric space... Maybe some sort of topology with the open sets being defined by like having a finite complement or something weird like that.

Insights appreciated!

Answer 1

You don't want the cofinite topology, but the cocountable topology (i.e., proper closed subsets are countable (including finite)).

Let $X$ be uncountable with the cocountable topology. Then the only convergent sequences are the eventually constant sequences, so there are no sequence in $Y$ that converges to $X-Y$, for all subsets $Y\subset X$. Now just choose $Y$ proper dense subset of $X$, e.g. $X-\{x\}$ for some $x\in X$.

Answer 2

Let $X=[0,1]^I$ where $I$ is uncountable (in the product topology, so a compact Hausdorff connected space) and let $Y$ be its subspace $$Y=\{(x_i)_{i \in I}\mid \{i: x_i \neq 0 \} \text{ is at most countable }\}$$

Then $Y$ is dense (show it intersects every basic open subset of $X$) but sequentially compact (in its subspace topology) and sequentially closed in $X$: every sequence of elements of $Y$ has also only countably many non-zero coordinates and so it "essentially lives" in a countable product of copies of $[0,1]$ and any limit of it also lies in it.

A very basic example (if you know some set theory) is $Y=\omega_1$ (the first uncountable ordinal, in the order topology) in $X=\omega_2$ (the ordinal of the next cardinality after it). These are ordinals of uncountable cofinality. Another classic example is $Y=\omega$ in $X=\beta\omega$, the Cech-Stone compactification of the natural numbers. Both of these might have been covered in whatever text or notes you're using.

Answer 3

Let $X=Y\cup \{\bar 0\}$ where $Y$ is the set of all functions from $\Bbb N$ to $(0,\infty),$ and $\bar 0$ is the constant function on $\Bbb N$ with $\bar 0(n)=0$ for all $n\in \Bbb N.$

Make $Y$ a discrete open subspace of $X.$ (So every subset of $Y$ is open in $X.$)

Define a local base for $\bar 0$ as $\{B(f):f\in Y\}$ where $B(f)= \{g\in X:\forall n\in \Bbb N\,(\,g(n)<f(n)\,)\}.$

It is easy to see that $Y$ is dense in $X.$

But $\bar 0$ is not the limit of a sequence in $Y.$ If $(g_n)_{n\in \Bbb N}$ is a sequence in $Y$ then for each $n\in \Bbb N$ let $f(n)=g_n(n)/2.$ Then $B(f)$ is an open nbhd of $\bar 0$ which is disjoint from $\{g_n:n\in \Bbb N\}.$