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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!

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    $\begingroup$ I'd think about uncountable ordinals. $\endgroup$ – Lord Shark the Unknown Jun 6 at 18:45
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    $\begingroup$ What about $\beta\omega$? It has no non-trivial convergent sequences, but it has proper subsets that are dense. $\endgroup$ – Jonathan Jun 6 at 19:00
  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet Jun 7 at 8:51
  • $\begingroup$ @Jonathan. Perfectly good example. But unlikely to be familiar to the proposer. $\endgroup$ – DanielWainfleet Jun 7 at 8:55
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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 $x\in X$.

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  • $\begingroup$ 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? $\endgroup$ – Mathematical Mushroom Jun 6 at 20:17
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    $\begingroup$ $U=X-\bigcup\{x_n\mid x_n\neq x\}$ 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 $\dots,x,x,x,\dots$. $\endgroup$ – user10354138 Jun 6 at 20:22
  • $\begingroup$ My space is compact and Hausdorff, yours is neither. It works though. $\endgroup$ – Henno Brandsma Jun 6 at 21:46
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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.

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

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

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