Solving Exercise 4 in Topology, Munkres.

The long line $L$ is the ordered set $S_\Omega \times [0, 1)$ in the dictionary order, with its smallest element deleted. And $S_\Omega$ denotes the minimal uncountable well-ordered set.

I want to prove that $L$ is normal, but not metrizable.

To show, my attempt was to show that $L$ is linear continuum. But it fails because $L$ is not linear continuum.

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    $\begingroup$ It is a linear continuum, though. $\endgroup$ – Henno Brandsma Dec 3 '19 at 7:01
  • $\begingroup$ Ah.. Could you give me naive explanation how can it satisfy intermediate property, which is one of conditions for linear continuum? $\endgroup$ – HooMun Dec 3 '19 at 7:48
  • $\begingroup$ Suppose $(\alpha,x) < (\beta,y)$ in $L$. If $\alpha <\beta+1$, consider $(\alpha+1,0)$ which lies inbetween. Other case $\beta=\alpha+1$, then find $x < x' < 1$. Otherwise $\alpha=\beta$ and $x < y$ and we can work in $[0,1)$. $\endgroup$ – Henno Brandsma Dec 3 '19 at 8:34

The long line is $S_\Omega \times [0,1)$ (not $(0,1]$) in the dictionary order topology (see p. 158 bottom (ex. 12) (all references to the 2nd edition of Munkres))

Then ex. 6 (chapter 3) on that same p. 158 says that if $X$ is well-ordered then $X \times [0,1)$ (in the dictionary order) is a linear continuum. So the long line is a linear continuum.

Finally ex. 8 on p. 206 (chapter 4) says that a linear continuum is normal.

So the long line is normal.

(Side note: it's not too hard but non-trivial, hence Munkres leaves it out of his text book for that reason I think, that any linearly ordered set in the order topology is even monotonically normal, which implies it is hereditarily normal (or completely normal), and this even extends to so-called generalised ordered spaces (which includes the lower limit topology as well). Now the result seems to depend on well-orderedness etc. while in fact it does not, really).

The long line is not metrisable, because it is limit point compact but not compact (for metrisable spaces these notions are equivalent). That's one of the possible arguments for it.

  • $\begingroup$ Always thanks a lot for your kind help! $\endgroup$ – HooMun Dec 3 '19 at 7:56

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