Topology Problems for Challenging Students

[Jamma]

no. try putting such a metric on the unit disk.

I'm sorry, I really don't understand.

What would be wrong with my metric on the unit disk? Maybe you misunderstood what I was saying- I define a new metric in terms of the older one. For any metric d, we can construct another one (called d^, say) which is defined as being equal to d except when d evaluates to a number larger than 1, in which case, you set d^ to return the value 1.

 

[lavinia]

I'm sorry, I really don't understand.

What would be wrong with my metric on the unit disk? Maybe you misunderstood what I was saying- I define a new metric in terms of the older one. For any metric d, we can construct another one (called d^, say) which is defined as being equal to d except when d evaluates to a number larger than 1, in which case, you set d^ to return the value 1.

The unit disk under the usual metric and your modification, is compact.

 

[Jamma]

The unit disk under the usual metric and your modification, is compact.

Oh, I think I see what you are getting at. I wasn't trying to imply that this procedure makes the space non-compact (the topology will be unaltered). All I'm saying is that for any non-compact complete metric space you can think of, if the metric isn't bounded then you can simply bound it in the way I described.