If $X$ is any space which is not second-countable, can one find a subspace $Y \subseteq X$ with $|Y| \leq \aleph_1$ which is also not second-countable? (Recall that a topological space $X$ is second-countable if it has a countable base.)

Note that if we require $|Y| \leq \aleph_0$ there are obvious counterexamples. For example, take $X$ to be the ordinal space $\omega_1$. Since $X$ has uncountably many isolated points, it is not second-countable, but every countable subspace is second-countable.

Looking through the non-second-countable spaces in π-base seems to suggest this is possible, but perhaps the spaces listed there are "too nice".

  • $\begingroup$ If you just want some bound on the size of $Y$, $|Y|\leq\mathfrak{c}^+$ works for rather trivial reasons (the $T_0$ quotient of any second-countable space has cardinality $\leq\mathfrak{c}$). $\endgroup$ – Eric Wofsey Apr 27 '16 at 19:06

The answer is yes.

There is special terminology and known results along these lines.

If every space with property $P$ has a subspace of size (at most) $\aleph_1$ with property $P$ then one says that property $P$ reflects in size $\le\aleph_1$ spaces.

For example non-metrizability (for compact spaces) reflects in size $\le\aleph_1$ spaces, i.e. every non-metrizable (compact Hausdorff) space has a non-metrizable subspace of size $\aleph_1$ (a result of Alan Dow). (Note that in this context $\aleph_1$ and $\omega_1$ are often used interchangeably, he writes $\omega_1$ in his paper. Similarly $\aleph_0$, $\omega$ and $\omega_0$ are used interchangeably.)

Recall that $w(X)$ denotes the weight of $X$, i.e. the smallest infinite cardinality of an open base for the topology of $X$.

The answer to your question is yes, see
Having a small weight is determined by the small subspaces,
Authors: A. Hajnal and I. Juhász
Journal: Proc. Amer. Math. Soc. 79 (1980), 657-658.
Abstract: We show that for every cardinal $ \kappa > \omega $ and an arbitrary topological space X if we have $ w(Y) < \kappa $ whenever $ Y \subset X$ and $ \vert Y\vert \leqslant \kappa $ then $ w(X) < \kappa $ as well. M. G. Tkačenko proved this for $ {T_3}$ spaces in [2]. We also prove an analogous statement for the $ \pi $-weight if $ \kappa $ is regular.

Here the property that "reflects" is "having weight $\ge\omega_1$".
If $w(X)\ge\omega_1$ then $w(Y)\ge\omega_1$ for some $Y\subseteq X$ with $|Y|\le\omega_1$. Indeed, if this were not the case, then taking $\kappa=\omega_1$ in the above theorem, we would have that every subspace $Y\subseteq X$ with $|Y|\le\omega_1$ satisfies $w(Y)<\omega_1$ (i.e. $w(Y)\le\omega_0$, $Y$ has countable weight, $Y$ is second-countable), and then the conclusion would be that $w(X)<\omega_1$ (i.e. $w(X)\le\omega_0$, $X$ has countable weight, $X$ is second-countable).

For reflection of non-metrizability (with a proof using elementary submodels) see:
An empty class of nonmetric spaces, Alan Dow,
Proc. Amer. Math. Soc. 104 (1988), 999-1001.

And here is a more-or-less random selection about reflecting other properties
(just a few sample links, out of many more).

On Dow's reflection theorem for metrizable spaces
Jerry E. Vaughan
TOPOLOGY PROCEEDINGS Volume 22, 1997, 351-360

Reflecting point-countable families, Zoltan T. Balogh
Proc. Amer. Math. Soc. 131 (2003), 1289-1296

Reflecting Lindelöf and converging $\omega_1$-sequences
Alan Dow, Klaas Pieter Hart
arXiv:1211.2764v2 [math.GN] http://arxiv.org/abs/1211.2764
(Submitted on 12 Nov 2012 (v1), last revised 17 Jul 2013 (this version, v2))

Reflecting Lindelöfness
James E. Baumgartner, Franklin D. Tall,
Topology and its Applications
Volume 122, Issues 1–2, 16 July 2002, Pages 35–49

Ideal reflections
Paul Gartside, Sina Greenwood, and David Mcintyre
Volume 27, No. 2, 2003, Pages 411-427


Here's a partial result. In particular, this answers your question in the affirmative for Hausdorff spaces assuming CH, or more generally for Hausdorff spaces if you replace $\aleph_1$ by $\mathfrak{c}$. (Or slightly more generally, you may replace "Hausdorff" with "no sequence has more than $\mathfrak{c}$ limits".)

Theorem: Let $X$ be a Hausdorff space with $|X|>\mathfrak{c}$. Then there is a subspace $Y\subset X$ of cardinality $\aleph_1$ which is not second-countable.

Proof: We choose a sequence $(x_\alpha)_{\alpha<\omega_1}$ of elements of $X$ by induction as follows: having chosen $x_\beta$ for all $\beta<\alpha$, choose $x_\alpha$ to be a point of $X$ which is not the limit of any sequence in $\{x_\beta\}_{\beta<\alpha}$. Such a point exists since there are only $\mathfrak{c}$ such sequences, and each has at most one limit.

Now let $Y=\{x_\alpha\}_{\alpha<\omega_1}$; suppose $Y$ is second-countable. By construction, the set $\{x_\beta\}_{\beta<\alpha}$ is sequentially closed in $Y$ for each $\alpha$, and hence closed in $Y$. But this means that for each $\alpha$, any basis for $Y$ has an element which contains $x_\alpha$ but does not contain $x_\beta$ for any $\beta<\alpha$. These sets must be distinct for different values of $\alpha$, so $Y$ cannot be second-countable.

  • $\begingroup$ Thank-you for the partial result. So I guess if one looks for a counterexample, it cannot be a Hausdorff space. (Actually, it's a bit stronger than that, since you're only using that convergent sequences have unique limits, which is a slightly weaker condition than Hausdorffness.) $\endgroup$ – Meta-мета-μετα-meta-мета-μετα May 3 '16 at 4:56
  • $\begingroup$ In fact, you only need that every convergent sequence has at most $\mathfrak{c}$ limits. So if you want to find a counterexample assuming CH, it must be a space which has a sequence with more than $\mathfrak{c}$ different limits. $\endgroup$ – Eric Wofsey May 3 '16 at 5:04

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