Topology - spaces, compactness

[edwinbrody]

Homework Statement

Find a metric space and a closed, bounded subset in it which is not compact.

Homework Equations

N/A

The Attempt at a Solution

I know that a metric space is a space that has definition such that a point has a distance to any other point. However, I know a compact space is bounded. I'm confused on what kind of subset of a metric space could be bounded yet not compact.

Background - I'm taking this course to fill a math minor, so my knowledge of abstract concepts is limited.

 

[edwinbrody]

Is there an interval or set that is neither closed nor open? I know if for example the rationals are being considered, they are neither closed nor open, and because of this fact, the rationals are not compact.

 

[losiu99]

Yes, there are sets which are neither closed nor open. Rationals are neither closed nor open regarded as a subset of reals, and indeed it's one of the things that prevents them from being compact.

As for the original question: rationals with usual distance, defined by the absolute value of the diference present fine example. We can take a closed interval, say, $$[0,1]$$, and because of incompleteness of rationals, it's not compact.

Compact space is necesarily bounded, but compactness enforces even stronger notion: total boundedness. Above example relies on incompleteness, but, for example, closed unit ball of infinite dimensional Banach space is closed, bounded and complete, yet still it's not compact.

 

[edwinbrody]

Yes, there are sets which are neither closed nor open. Rationals are neither closed nor open regarded as a subset of reals, and indeed it's one of the things that prevents them from being compact.

As for the original question: rationals with usual distance, defined by the absolute value of the diference present fine example. We can take a closed interval, say, $$[0,1]$$, and because of incompleteness of rationals, it's not compact.

Compact space is necesarily bounded, but compactness enforces even stronger notion: total boundedness. Above example relies on incompleteness, but, for example, closed unit ball of infinite dimensional Banach space is closed, bounded and complete, yet still it's not compact.

So, for instance in this example, If I take the space bounded by a circle of radius 2 centered on the origin, and the square bounded by x from [0,1] and y from [0,1] as the subset within it, the subset is closed and bounded, but because of incompleteness of the rationals this subset is not compact?

 

[losiu99]

Yes. It's enough to take just $$[0,1]$$ as a subset of rationals, as it is closed and bounded, yet not compact.