1. ## Topology 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?

2. You posting has some missing symbols.
What are they?

3. Are you talking about the s.t. ? I mean "such that"...sorry

4. Is (M, {∅,M}) hausdorff?

5. Originally Posted by JohnStaphin
Is (M, {∅,M}) hausdorff?
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?

6. 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

7. Originally Posted by JohnStaphin
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...
You have to be a little careful. What happens if M is the empty of singleton set? However, if M is not empty or singleton then there are two points $\displaystyle x,y \in M$ with $\displaystyle x\not = y$. And then it is not Hausdorff.

8. 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?