Are There Subsets of the Real Line That Are Compact But Not Connected?

[Salamon]

Let's take the real line.

I understand that there exists a subset of the real line which is connected and compact. Ex: [0,1]
I understand that there exists a subset of the real line which is neither connected or compact.
Ex: (0,1) U (5,6)

Do there exist any subsets of the real line which are compact and not connected?
Do there exist any subsets of the real line which are connected and not compact?

 

[R136a1]

Do there exist any subsets of the real line which are compact and not connected?
Do there exist any subsets of the real line which are connected and not compact?

Yes to both. Try to find counterexamples.

 

[Salamon]

I know all closed intervals are compact. I am not sure why this is so.

I am assuming [0,1] U [2,3] is compact but not connected?
And (0,1) is not compact but is connected,right?

 

[R136a1]

I know all closed intervals are compact. I am not sure why this is so.

What is your definition of compact?

I am assuming [0,1] U [2,3] is compact but not connected?
And (0,1) is not compact but is connected,right?

Right!

 

[Salamon]

A set or space is compact iff every open cover of the space has a finite subcover.

Is everything that I said still true if the real line is a topological space which is topologized with the usual topology?

 

[R136a1]

A set or space is compact iff every open cover of the space has a finite subcover.

See http://planetmath.org/proofofheineboreltheorem under the heading "The case n=1, the closed interval is compact". This contains an elementary proof of the fact that closed intervals in $\mathbb{R}$ are compact.
If you already know the Heine-Borel theorem however, then you know that compactness in $\mathbb{R}$ is equivalent to closed and bounded, and the result that closed intervals are compact becomes a lot easier to prove.

Is everything that I said still true if the real line is a topological space which is topologized with the usual topology?

Yes.

 

[HallsofIvy]

It is fairly easily provable that every compact set is closed.
It is fairly easily provable that every compact set, in a metric space so that "bounded" is defined, is bounded.

Much harder is the proof that, in the real numbers with the "usual" topology, sets that are both closed and bounded are compact. Your statement that "all closed intervals are compact" is NOT true. [0, \infty) is closed but NOT compact.

 

[WWGD]

It is fairly easily provable that every compact set is closed.
It is fairly easily provable that every compact set, in a metric space so that "bounded" is defined, is bounded.

Much harder is the proof that, in the real numbers with the "usual" topology, sets that are both closed and bounded are compact. Your statement that "all closed intervals are compact" is NOT true. [0, \infty) is closed but NOT compact.

Maybe you mean every compact set is closed in metric topologies? Note that compact sets are closed under Hausdorff topologies, but not necessarily otherwise.

Still, notice that while compact implies closed in Hausdorff spaces, the opposite does not always hold; take, e.g., the interval $[a,b] \cap \mathbb Q$. This is closed in $\mathbb Q$, but not compact there, because you can find sequences there without convergent subsequences ( which cannot happen in a compact subset of a metric space ), like, e.g., the sequence 1, 1.4, 1.414,... that "should converge" to $\sqrt 2$