# 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

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!

• I'd think about uncountable ordinals. – Lord Shark the Unknown Jun 6 at 18:45
• What about $\beta\omega$? It has no non-trivial convergent sequences, but it has proper subsets that are dense. – Jonathan Jun 6 at 19:00
• The tightness of a space is the least infinite cardinal $k$ such that whenever $Y\subset X$ and $p\in \bar Y,$ there exists $Z\subset Y$ with $p\in \bar Z,$ and with the cardinal of $Z$ being at most $k$.... Metric spaces have tightness $\aleph_0$. ("Countably tight".) There are many "well-known" spaces with uncountable tightness, and they are all examples for your Q. – DanielWainfleet Jun 7 at 8:51
• @Jonathan. Perfectly good example. But unlikely to be familiar to the proposer. – DanielWainfleet Jun 7 at 8:55

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 $$x\in X$$.

• Thanks. Can you explain to me why the only sequences are the eventually constant ones? Say $x_n \rightarrow x$ and let $U$ be a neighborhood of $x$. Then $\exists M \in \mathbb{N}$ s.t. $\forall n > M$ we have $x_n \in U$... Okay, then what? – Mathematical Mushroom Jun 6 at 20:17
• $U=X-\bigcup\{x_n\mid x_n\neq x\}$ is an open neighbourhood of $x$, so must contain eventually every term of the sequence. But it doesn't contain any term of the sequence that is not $x$, so eventually the sequence is $\dots,x,x,x,\dots$. – user10354138 Jun 6 at 20:22
• My space is compact and Hausdorff, yours is neither. It works though. – Henno Brandsma Jun 6 at 21:46

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.

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)

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 $$\{g_n:n\in \Bbb N\}.$$

• I wanted to give an elementary example that does not need any background in Set Theory. – DanielWainfleet Jun 7 at 2:33