Theorem: If a topological space $(X, \tau)$ has a countable base (2nd countable), then for every $Y \subseteq X$ its closure $cl(Y)$ is the set of limit points of sequences $(x_j)_{j\in \mathbb N}$ where $x_j \in Y$ for all $j \in \mathbb N$.

Now I wanted to find a "non-example" for this Theorem, namely a not second countable space with a subset $Y$ who has a limit point that could not be expressed as the limit point of a sequence in $Y$.

i) For an uncountable $X$, the co-finite topology $$ \tau = \{ A \subseteq X : A = \emptyset \mbox{ or } A^C \mbox{ is finite } \} $$ is not second-countable. In this topology each finite set has no limit points, and each infinite set is dense, i.e. every point of $X$ is a limit point. So if $Y \subseteq X$ is infinte we can select some $\{ x_i \} \subseteq Y$ which is infnite, and this sequence will converge to every point of $X$, i.e. to each limit point of $Y$. For if $x \in X$ and $U$ is some open set containing $x$, then $U^C$ is finite, and so for some $N$ all $x_i$ with $i > N$ must lie in $U$. So this example is not working.

ii) The discrete topology for uncountable $X$ is not second countable, but because each point is isolated it has no limit points.

So these two "non-examples" would not work, do you know a not second countable space who has a subset containing a limit point which is not the limit point of a sequence in that set?

  • 1
    $\begingroup$ Try $[0,\omega_1], \omega_1$ being the first uncountable ordinal. No sequence in $[0,\omega_1)$ can converge to $\omega_1$. $\endgroup$ – Kevin Carlson Jan 6 '14 at 21:27

Spaces where the limit points of a set $A$ are also limits of sequences in $A$ are called Fréchet-Urysohn spaces. Note that every first-countable space is a Fréchet-Urysohn space.

One example is the mapping cylinder of the inclusion $i:(0,1]\hookrightarrow [0,1]=I$, i.e. the space $(0,1]\times I\sqcup I×\{0\}$ with $(a,0)$ identified to $a\in I$. Here an open neighborhood of $(0,0)$ is a set $U=\{(0,0)\}\cup U'$ where $U'\subseteq (0,1]×I$ is open and contains $(0,\epsilon)×\{0\}$ for some $ϵ>0$. It can be shown that no sequence in $(0,1]×(0,1]$ can converge to $(0,0)$ although $(0,0)$ is in the closure of that subset.

Another example is $X=(\Bbb R,\tau)$ where $\tau$ is the cocountable topology. Here any uncountable set $A$ is dense, but a convergent sequence is eventually constant (!), so it can never reach something outside of $A.$

The difference between the two examples is that the first space is still a sequential space, that is a space where a subset $A$ is closed if it contains the limits of all sequences in $A$. Being a sequential space is necessary for being a Fréchet-Urysohn space. In the second space each set trivially contains the limits of its sequences, but not each set is closed, so this is not a sequential space.


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