4

I've just started working through a topology textbook, and I'm sure I'm being silly, but I can't for the life of me make any headway on the following question:

Let $(M,\rho)$ be a pseudometric space, $(M^*,\rho^*)$ its metric identification, $h(x)$ the canonical projection map from $M\to M^*$.

Then a set $A$ in $M$ is closed (open) iff $h(A)$ is closed (open) in $M^*$.

I have $A$ in $M$ open $\implies$ $h(A)$ open in $M^*$.

The thing that's tripping me up is considering say:

$\{1,2\}$ with the trivial pseudometric. The metric identification is $\{[1]\}$, so $\{[1]\}$ is open in the metric identification, but $\{1\}$ is not open in the pseudometric space, since every disk around $1$ contains $2$ as well.

Thanks I'm sure I'm probably doing something stupid, but after googling it I couldn't find an explanation so I came here.

jgon
  • 28,469
  • 2
    This mistake is in the exercise 2C.3 on page 20 of the book General Topology (1970) by Stephen Willard. – beroal Jan 24 '18 at 10:32

1 Answers1

5

The characterization of the topology on $M$ that you have written (which, I see, is the characterization on the wikipedia page for pseudometrics) has an error.

It should be: "a set $A$ in $M$ is closed (open) iff $A$ is saturated and $h(A)$ is closed (open) in $M^*$".

Lee Mosher
  • 120,280
  • I think I know what you mean, but what is the definition of saturated?

    Yeah I checked wiki after I read this in my textbook and was kind of surprised to see it there too.

     – jgon Aug 19 '14 at 01:18
  • 1
    A subset $A \subset M$ is saturated if for every $x,y \in M$, if $x \in A$ and if $x,y$ have the same image in $M^$ then $y \in A$. Another way to say it is that a saturated subset of $M$ is the same thing as the preimage of a subset of $M^$ under the canonical projection. – Lee Mosher Aug 19 '14 at 01:21
  • Awesome that's sorta what I expected it to be thanks. – jgon Aug 19 '14 at 01:22