《CCTV空中剧院》 20171016 越剧《五女拜寿》 12

Question

Let $(X, \tau_1)$ be a Hausdorff second countable topological space and $(X, \tau_2)$ be a space with a coarser topology, but still Hausdorff. Is $\tau_2$ still second countable?

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The question was closed as I was asked to give more background as to what I am trying to do and why the question is important... the issue is that it's a piece in a much bigger puzzle. It's kind of a pity it was closed because it was getting answers with interesting insights.

At any rate, here it goes. I am trying to find a minimal set of physical assumptions that I can encode into math and find the most general but still physically meaningful space for states. Therefore I need to go back and forth trying to understand what initial assumptions lead to what.

In particular, I know that the space of statistical has to have at least the following features:

• it has to be a $\mathsf{T}_0$ second countable topological space (from experimental verifiability)
• it is a convex space such that the convex operation is continuous (from the ability to perform mixtures
• it has to allow an entropy function (to characterize the variability of the elements within the ensemble)

What I am trying to understand, overall, is whether the premises (which are better specified in our draft) are enough to show that we embed continuously into a Hausdorff second countable locally convex topological vector space (which would be automatically be metrizable, as far as I understand). I can show the embedding into a vector space. I have been struggling with the topology for months. I am also discussing this with other mathematicians to understand how to proceed.

When I asked this question, I was specifically trying to get another topology from statistical quantities, which would simply be continuous affine function over the space of ensembles. I can show that these induce semi-norms on both the ensemble space and the ambient vector space, so I would have automatically a topology on both, that would make, as far as I understand, the ambient vector space a Hausdorff locally convex TVS. So, if second countability were inherited by the topology on the ensemble space, I may have been able to proceed further.

I also have the problem that the entropy does give me a topology (that at least in classical and quantum mechanics should correspond to the topology given by the inner product). But I do not know if I can prove that the topology of the space is exactly that topology. So, knowing relationship between coarser and finer topologies would be very useful. For example, if I could know when I have a/the finest second countable topology.

I understand this is all very vague... I have many piece on the table and trying to understand how to make them fit in the best way... which is why I narrowed the question to something very specific without context. In general, relationships between finer/coarser topologies is what I was trying to find... much like for the subset topology.

Answer 1

Quite often, it won't be. Let $\tau_1$ be the norm topology of a separable Hilbert space $X$, and $\tau_2$ the weak topology. Then, if $X$ is not finite-dimensional, $\tau_2$ is not even first-countable, a fortiori it's not second-countable.

Answer 2

Let $\tau_1$ be the discrete topology on $\mathbb N$ and let $\tau_2$ be any non-first-countable Hausdorff topology on $\mathbb N$, for example, the collection of all sets $U\subseteq\mathbb N$ such that either $1\notin U$ or else $U$ has asymptotic density $1$.

Answer 3

Q 1. For which second-countable Hausdorff spaces $X$, any coarser Hausdorff topology is second-countable?

If we additionally assume regularity we get from Urysohn metrization theorem:

Q 2. For which separable metrizable spaces $X$, any coarser Hausdorff topology is second-countable?

A class of spaces which answers Q 1 are second-countable minimal Hausdorff spaces. A non-regular example was constructed in Minimal topological spaces by Berri in remark 1.5.

A class of spaces for which Q 1 fails are second-countable Hausdorff spaces which admit a discrete family $\mathcal{U} = \{U_n : n\in\mathbb{N}\}$ of non-empty open sets. Indeed, take $x_n\in U_n$ for each $n$ and let $N = \{x_n : n\in\mathbb{N}\}$. Then $N$ is discrete and closed. Take a Hausdorff topology $\sigma$ on $N$ which is not first countable at $x_m\in N$ (see here for some examples). Let $\tau'$ be topology on $X$. Declare $\tau = \{U\subseteq X : U\in \tau' \land U\cap N\in \sigma\}$. Then $\tau$ is a Hausdorff topology on $X$ which is coarser than the original topology $\tau'$, and it's not first countable at $x_m\in N$. Proof:

1. $\tau$ is a topology:

Clearly $\emptyset \in \tau'$ and $\emptyset\cap N = \emptyset\in \sigma$, so $\sigma\in \tau$. Also $X\in \tau'$ and $X\cap N = N\in \sigma$ so $X\in \tau$. If $V_1, V_2\in \tau$, then $V_1\cap V_2\in \tau'$ and $V_1\cap V_2\cap N = (V_1\cap N)\cap (V_2\cap N)\in \sigma$ so $V_1\cap V_2\in \tau$. If $V_i\in \tau$, then $\bigcup_{i\in I} V_i\in \tau'$ and $(\bigcup_{i\in I} V_i)\cap N = \bigcup_{i\in I} (V_i\cap N) \in \sigma$, so $\bigcup_{i\in I} V_i\in \tau$.

2. $(X, \tau)$ is Hausdorff:

If $x, y\in X\setminus N$ take disjoint $V_1, V_2\in \tau'$ with $x\in V_1, y\in V_2$ and $V_1, V_2\subseteq X\setminus N$. Then $V_1\cap N = V_2\cap N = \emptyset \in \sigma$ so $V_1, V_2\in \tau$. If $x, y\in N$ take disjoint $A_1, A_2\in \sigma$ with $x\in A_1, y\in A_2$ and let $V_k = \text{St}(A_k, \mathcal{U}) = \bigcup\{U\in\mathcal{U} : U\cap A_k\neq\emptyset\}$. Then $V_k\in \tau'$ and $V_k\cap N = A_k\in \sigma$ so $V_1, V_2\in \tau$, and $V_1\cap V_2 = \emptyset$. If $x\in X\setminus N$ and $y\in N$, take $x\in V\in \tau'$ with $V\cap U\neq\emptyset$ for at most one $U\in\mathcal{U}$. If $V\cap U = \emptyset$ for all $U\in \mathcal{U}$, take $V_1 = V$ and $V_2 = \bigcup\mathcal{U}$ and note that $V_1, V_2\in \tau'$ and $V_1\cap N = \emptyset\in \sigma, V_1\cap N = N\in \sigma$ so $V_1, V_2\in \tau$. If $V\cap U \neq \emptyset$ for some $U$ and $y\notin U$, then take $V_1 = V$ and $V_2 = \bigcup\mathcal{U}\setminus U$, then $V_1, V_2\in \tau'$ and $V_1\cap N = \emptyset\in \sigma$ and $V_2\cap N = N\setminus \{x_n\}\in \sigma$ for some $x_n\in N\setminus \{y\}$, so $V_1, V_2\in \tau$ and also $V_1\cap V_2 = \emptyset$. If $V\cap U\neq\emptyset$ for some $U$ and $y\in U$, then take disjoint $W_1, W_2\in \tau'$ with $x\in W_1, y\in W_2$ and $W_1\subseteq V$ and $W_2\subseteq U$ and let $V_1 = W_1$ and $V_2 = \bigcup \mathcal{U}\setminus U \cup W_2$, then $V_1, V_2\in \tau'$ and $V_1\cap N = \emptyset\in\sigma$ and $V_2\cap N = N\in \sigma$ so $V_1, V_2\in \tau$ and also $V_1\cap V_2 = \emptyset$.

3. $\tau$ is coarser than $\tau'$:

If $U\in \tau$ then $U\in \tau'$ by definition.

4. $(X, \tau)$ is not first countable at $x_m$:

Note that $(N, \sigma)$ is a subspace of $(X, \tau)$ since if $U\in \tau$ then $U\cap N\in\sigma$, and if $A\in \sigma$ then $V = \text{St}(A, \mathcal{U}) = \bigcup\{U\in\mathcal{U} : U\cap A\neq\emptyset\}$ is such that $V\cap N = A$ and $V\in \tau$. If $(X, \tau)$ were first countable at $x_m$, then $(N, \sigma)$ would be first countable at $\sigma$, contradiction.

In particular, the spaces for which Q 2 is true are precisely compact metrizable spaces.

If $X$ is compact metrizable, then $X$ is separable metrizable, and it's also minimal Hausdorff so that there's no strictly coarser Hausdorff topology. And if $X$ is non-compact and metrizable, then we can find a metric $d$ on $X$ which is not totally bounded, and so we can find an infinite $\varepsilon$-net on $X$, and so a discrete family of non-empty open sets by taking open balls at points of this net of a fixed small enough radius, and apply previous construction. Alternatively one can take infinite closed discrete set directly, define topology on $X$ just as previously, and use normality of $X$ to show that the topology is Hausdorff.