Given the topological space (M, {∅,M}), prove that there isn't a metric "d" on M, s.t. (M, {∅,M}) is the topological space arising from (M,d)...

Any suggestions?

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- Sep 17th 2008, 04:56 PMJohnStaphinTopology question
Given the topological space (M, {∅,M}), prove that there isn't a metric "d" on M, s.t. (M, {∅,M}) is the topological space arising from (M,d)...

Any suggestions? - Sep 17th 2008, 05:07 PMPlato
You posting has some missing symbols.

What are they? - Sep 17th 2008, 05:10 PMJohnStaphin
Are you talking about the s.t. ? I mean "such that"...sorry

- Sep 18th 2008, 08:14 AMJohnStaphin
Is (M, {∅,M}) hausdorff?

- Sep 18th 2008, 08:20 AMThePerfectHacker
Do you know the definition of being Hausdorff? It means for any two points in $\displaystyle M$ we can find two open sets in the topological space so that they contain these points and are disjoint. Is it possible to do it here? Note, the topological space is simply $\displaystyle \{\emptyset, M\}$. So using only these two is it possible to have a Hausdorff property?

- Sep 18th 2008, 09:02 AMJohnStaphin
Of course not..Since (M, {∅,M}) only allows you to construct sets in terms of M, and ∅, you could actually never find a disjoint open set that contains an element of M and not all other elements as well...

So the proof goes something like..Every Metric space is hausdorff, And from observing that (M, {∅,M}) isn't, then this topological space couldn't have came from (M,d)

..something like that anyway - Sep 18th 2008, 09:09 AMThePerfectHacker
- Sep 18th 2008, 09:29 AMJohnStaphin
Well if M is a singleton set...or just a finite number of points even..then under (M,d) it will be closed...Now, if M is the empty set...I have no idea...lol.. Is it Hausdorff? It seems like you could say what ever you wanted to say...There aren't point to choose in ∅, to see if it is hausdorff...so does that say it is or isnt?