Is a single point in R compact?

[fraggle]

Is a single point in R compact?

It seems obvious since every open cover of a single point in R can clearly have a finite subcover.

However, I have a little uncertainty (i.e possible convention that says otherwise?) so just wanted to check before using it in a proof.
thanks

 

[VeeEight]

A subset of R is compact if it is closed and bounded (Heine Borel). A set consisting of a single point is certainly bounded and closed and therefore compact

 

[Office_Shredder]

It seems obvious since every open cover of a single point in R can clearly have a finite subcover.

That's it. No more work needed

A subset of R is compact if it is closed and bounded (Heine Borel). A set consisting of a single point is certainly bounded and closed and therefore compact

I would probably fail anyone using this argument on principle

 

[HallsofIvy]

Technical point: it makes no sense to talk about a point being compact. What you mean is that a set containing a single point (a "singleton" set) is compact. That's true in any topology, not just R or even just in a metric space. Given any open cover for {a}, there exist at least one set in the cover that contains a and that set alone is a "finite subcover".

 

[g_edgar]

I would probably fail anyone using this argument on principle

And I would say you would be in error to do so.

What would you say to this:

Let p be the point. We know that there is some compact space K. The map defined by f(x) = p for all x \in K is continuous. The continuous image of a compact set is compact. Therefore \{p\} is compact. QED

 

[Office_Shredder]

And I would say you would be in error to do so.

I was mostly being humorous. However, I know a lot of teachers that specifically don't want people to use Heine-Borel to prove the compactness of sets because it misses the point (which is to demonstrate your knowledge of what compactness means).

 

[HallsofIvy]

And I would say you would be in error to do so.

What would you say to this:

Let p be the point. We know that there is some compact space K. The map defined by f(x) = p for all x \in K is continuous. The continuous image of a compact set is compact. Therefore \{p\} is compact. QED

Oh, surely you can find an even more complicated proof than that!