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If X is a Hausdorff space and Y is a dense subspace of X, can the density of Y exceed the density of X? The density of a space X is the least infinite cardinal C such that X has a dense set of cardinal C or less.

DanielWainfleet
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2 Answers2

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See the following paper:

R. Levy, R.H. McDowell, Dense subsets of X.
Proc. Amer. Math. Soc. 50 (1975), 426-430.

http://www.ams.org/journals/proc/1975-050-01/S0002-9939-1975-0370506-8/S0002-9939-1975-0370506-8.pdf

From the introduction: "... use this result to obtain a nonseparable space $X$ such that $\beta X$ is separable."

Since $X$ is dense in $\beta X$, the above result shows that you could have a separable space, namely $\beta X$, and a non-separable dense subset, namely $X$.

I got to this reference after I got first to a paper by J Talponen - ‎2015,
http://arxiv.org/pdf/1506.06080.pdf
when I googled:
increasing density of dense subsets
and looked at the first couple of pages.

Levy and McDowell also describe a more standard example, 5.3.
Let $X$ be the product of continuum many copies of the closed unit interval $[0,1]$. Let $Y$ be the subspace of $X$ of functions $f$ that take value $0$ at all but finitely many coordinates. (I think this is known as $\sigma$-product, may try to find another reference ... enclosed some links in a comment. $X$ is separable by the Hewitt-Marczewski-Pondiczery theorem.) Then $Y$ is dense in $X$, but is not separable.

Mirko
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    If for example you take $X = [0,1]^{[0,1]}$ then you can show directly that it is separable, without HMP. Take for example the set of polynomials with rational coefficients, truncated so as to have range in $[0,1]$. Then, knowing what we do about the product topology, it's easy to show that any basic open set contains such a polynomial. – Nate Eldredge Jul 25 '15 at 16:35
  • One possible reference about $\sigma$-products and $\Sigma$-products is Encyclopedia of General Topology, ed. by K.P. Hart, Jun-iti Nagata, J.E. Vaughan https://books.google.com/books?id=JWyoCRkLFAkC&pg=PA40&lpg=PA40#v=onepage&q&f=false see section b-3 there. They were intorduced by Corson in 1959 http://www.jstor.org/stable/2372929 . See also Dan Ma's blog https://dantopology.wordpress.com/tag/sigma-products/ – Mirko Jul 25 '15 at 17:07
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Partial answer:

In general $d(X)\leq d(Y)$ for dense subsets. So if the answer were "no", then we'd always have $d(Y)=d(X)$, which in this case is equivalent to saying that density is a monotone cardinal function. Now Hodel's article in the Handbook of Set-theoretic topology explicitly states that density is not monotone, but that it is monotone for open subsets, so I guess that the answer is "no" in case your $Y$ is open, but it looks like there does indeed exist examples of non-open $Y$ with $d(Y)>d(X)$, but I am not sure if they can be chosen to be dense.

Damian Reding
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