# Lower limit topology doesn't arise from metric; proof by contradiction.

This problem is in Tao's Epsilon of the room, 1.6.11.

Show that the half-open topology (Example 1.6.22) is Hausdorff, but does not arise from a metric. (Hint: assume for contradiction that the half-open topology did arise from a metric; then show that for every real number x there exists a rational number q and a positive integer n such that the ball of radius 1/n centred at q has infimum x.)

And I proved it in below way; I think this proof has some missing points; Could you verify this proof?

Clearly, Let $F$ be half-open interval topology and $x,y \in \mathbb{R}$ and without loss of generality, $x<y$. Then $[x,y)$ and $[y,y+1)$ is clearly open neighborhood of $x,y$ without nonempty intersection. Hence this topology is hausdorff. Now we want to show that it doesn't arise from metric space. For contradiction, suppose it is arise from metric space. By definition of metrizable topology, for any $[x,y) \in F$ can be represented as union of open balls $B(x_0, r) = \{ x \in X : d(x_0 , x) < r\}$. We know that $\mathbb{Q}$ is dense in $\mathbb{R}$, so for any open sets $[a,b) \in F$, $\mathbb{Q} \cap [a,b) \neq \emptyset$. Hence $F$ is separable. With the fact that $\mathbb{Q} \cap [a,b)$ is dense in $F$ and $F$ is metrizable, there exists $x \in \mathbb{R}$ such that $x= \inf\{B(q,\frac{1}{n})\}$. However this is contradiction when $x \notin \mathbb{Q}$ since $\inf\{B(q,\frac{1}{n})\} = (q-\frac{1}{n},q+\frac{1}{n})$ has infimum as $q-\frac{1}{n}$, which is rational number. Therefore, $F$ is not metrizable.

-

One small problem occurs at the end, when you say

$\inf \{ B ( q , \frac 1n ) \} = ( q - \frac 1n , q + \frac 1n )$

Apart from a notational difficulty (a real number is not an interval), you seem to be assuming that the balls arising from the metric generating the half-open interval topology are open intervals in $\mathbb{R}$. Note that if this were the case, then the half-open interval topology would be coarser than the usual topology, which is not so.

Note the following: $X$ is a metric space, and $D \subseteq X$ is dense, one can show that the family $$\mathcal{B} = \{ B_{1/n} ( x ) : x \in D , n \in \mathbb{N} \}$$ is a base for the topology.

Use the above with your observed fact that $\mathbb{Q}$ is dense in $F$ to prove Tao's hint (pick an appropriate open neighbourhood of $x$). Next observe that if $x,y \in \mathbb{R}$ are distinct and $p,q \in \mathbb{Q}$, $n,m \in \mathbb{N}$ are such that $$x = \inf ( B_{1/m} ( p ) ) \quad y = \inf ( B_{1/n} ( q ) )$$ then either $n \neq m$ or $p \neq q$. From here we can set up an injective function $\mathbb{R} \to \mathbb{Q} \times \mathbb{N}$, which is problematic.

As an aside, you actually note something that is very important in your run-down (emphasis added):

Hence $F$ is separable.

One basic fact about metric (metrizable) spaces is that they are separable iff they are second-countable. This is the usual route (in my humble opinion) to show that $F$ is not metrizable (and, actually, Tao's hint gets you pretty close to going this way).

-
Thank you for detailed explanation! –  user124697 Feb 19 '14 at 8:17

Another proof to see it is not metrizable: $X$ is separable: every set $[a,b)$ with $a < b$ contains a rational (already in $(a,b)$), so $\mathbb{Q}$ is dense and countable. But $X$ is not second countable: let $\{ B_i: i \in I\}$ be a base for $X$. For each $x$ in $X$, we pick $i(x) \in I$ such that $x \in B_{i(x)} \subset [x, x+1)$, by the definition of being a base.

But note that if $x < y$, say, then $x \notin B_{i(y)}$, as $B_{i(y)} \subset [y, y+1)$, which does not contain $x$, and $x \in B_i(x)$ by construction, hence $B_{i(x)} \neq B_{i(y)}$ for two distinct $x,y$ in $X$. So the base has at least as many elements as $X$ has, which is certainly uncountable...

For a metrizable space, being separable implies having a countable base, so this shows that $X$ is not metrizable.

-
Wow, this is simple proof. Actually I check Arthur's answer because it is the explanation of what Tao's hint mean, but I also thank you for another way to prove this. –  user124697 Feb 19 '14 at 8:29
@user124697 you're welcome, this is the proof I have been taught originally for this. It's what I think most topologists would give as a proof (it actually shows a bit more, namely that the weight of $X$ is continuum). Note that both proofs use the order in some way, though my proof does it in an easier way (no sups or infs) –  Henno Brandsma Feb 19 '14 at 8:37
Very nice, it's one of those proofs that seem so obvious after you see it. But I struggled with this problem for quite a while. –  Gregory Grant Apr 3 at 13:43