I Problem with countably compact space

[facenian]

Helo. A problem in TOPOLOGY by Munkres states that for a $T_1$ space $X$ countable compactness is equivalent to limit point compactes(somtimes also known as Frechet compactness). Countable compactness means that every contable open covering contains a finite subcollection that covers $X$.
He gives a Hint to show that a limit point compact and $T_1$ space is countably compact. However he does not give any clue for the reciprocal proposition.
My question is: Is the reciprocal poposition correct or the use of the word equivalent is a typo?

 

[fresh_42]

I've found the following example on Wikipedia: (I haven't checked it in detail, but it looks promising.)

A compact Hausdorff space that is not sequentially compact
The set $\{0,1 \}$ provided with the discrete topology is compact. Therefore, according to the theorem of Tychonoff also the set $\{0,1 \}^{[0,1]}$ all functions on the interval $[0,1]$ to $\{0,1 \}$ provided with the product topology is also compact, and it is Hausdorff.

The product topology on $\{0,1 \}^{[0,1]}$ means that a sequence of functions converges if it converges pointwise.

But this space is not sequentially compact.

A sequence of functions that does not contain a convergent subsequence can be defined as follows:

The dual places of a real number are an infinite sequence of zeros and ones.
The sequence $(a_{n})_{n \in \mathbb{N}} \in \{0,1 \}^{[0,1]}$ is now defined as follows:
$a_{n}(x)$ is the $n-$th dual digit of the number $x$.
To a subsequence $(a_{n_{k}})_{n_k \in \mathbb{N}}$ we define a number $y$ as follows. We set in the binary representation of $y$ at $n_k$ a $0$ if $k$ is even, a $1$ if $k$ is odd, and $0$ at all other locations. That means the sequence $(a_{n_{k}})_{n_k \in \mathbb{N}}$ does not converge, because at $y$ the values jump back and forth. Thus the sequence $(a_{n})_{n \in \mathbb{N}}$ can not have a convergent subsequence.

The English Wikipedia has:
In a sequential (Hausdorff) space sequential compactness is equivalent to countable compactness.

 

[facenian]

Since limit point compactness-----> sequential compactness. And compact(Hausdorff) -----> countably compact(T_1) . This example is interesting however it is not a counter example to the use of the word equivalent in the proposed exercise.

 

[fresh_42]

As I see it, limit point compactness and sequential compactness is the same, and only due to a different translation. Now Hausdorff implies $T_1$. Thus the example shows $\text{ countably compact } T_1 \nRightarrow \text{ limit point compactness }$. In order to be true, the Hausdorff space has also to be a sequential space. Without it, the equivalence doesn't hold.

Am I wrong here?

 

[member 587159]

Helo. A problem in TOPOLOGY by Munkres states that for a $T_1$ space $X$ countable compactness is equivalent to limit point compactes(somtimes also known as Frechet compactness). Countable compactness means that every contable open covering contains a finite subcollection that covers $X$.
He gives a Hint to show that a limit point compact and $T_1$ space is countably compact. However he does not give any clue for the reciprocal proposition.
My question is: Is the reciprocal poposition correct or the use of the word equivalent is a typo?

The statement is true. I will provide a proof for "countably compact => limit point compact. In fact, the proof is quite straightforward: By contraposition, we start with a countable infinite set without limit point, and we find a countable open cover without finite subcover.

Proof: Let $(X, \mathcal{T})$ be a topological space. Suppose that there is a countable infinite subset $S$ without limit point.

This means that $S$ must be closed (a set without limit points contains all its limit points, and hence is closed), but also that for any $x \in S$ there exists an open neighborhood $V_x$ of $x$ such that $V_x \cap (S \setminus \{x\}) = \emptyset$.

The collection $\{V_x \mid x \in S\} \cup \{X \setminus S\}$ is a countable open cover without finite subcover, as it is not possible to leave out $V_y$ for a certain $y \in S$. This ends the proof.

Note that I didn't use the $T_1$ property, but this is required for the other direction.

 

[member 587159]

As I see it, limit point compactness and sequential compactness is the same, and only due to a different translation. How Hausdorff implies $T_1$. Thus the example shows $\text{ countably compact } T_1 \nRightarrow \text{ limit point compactness }$. In order to be true, the Hausdorff space has also to be a sequential space. Without it, the equivalence doesn't hold.

Am I wrong here?

Yes, I think you are. I didn't look at the example though. The statement (in one direction) doesn't even seem to require $T_1$ (or stronger things like Hausdorff).

 

[fresh_42]

Yes, I think you are. I didn't look at the example though.

Me neither, just found it. Whether $T_1$ is used or not depends on Tychonoff, so that isn't an obvious criterion. As soon as we go into the details of the zoo, things become less intuitive quickly. But still an interesting puzzle: How exactly are the inclusions between the two definitions of compactness, which direction needs what, how do sequential Hausdorff spaces come into play, where are they settled in this environment between $T_0,T_1,T_2$ and compactness, and last but not least: what does the example show and what doesn't.

I often read "in metric spaces ..." and then what follows is worthless for topology in general. I admit, I'm not the biggest fan of its zoo, so I'm not sure whether I want to solve the puzzle. Nevertheless, it's an interesting one.

P.S.: And it gets even messier if we bring in the term Fréchet compactness - just for the sake of confusion.

 

[facenian]

As I see it, limit point compactness and sequential compactness is the same, and only due to a different translation.
Am I wrong here?

They are the same in a metric space. Though sequential compactness --------> limit point compactness in general(always) the reciprocal needs 1st countability.

 

[Erland]

How does Munkres define "limit point compact"?

 

[facenian]

How does Munkres define "limit point compact"?

A space X is said to be limit point compact if every infinite subset of X has a limit point.
The thread's question was answered in post #5.

 

[member 587159]

A space X is said to be limit point compact if every infinite subset of X has a limit point.
The thread's question was answered in post #5.

Note that that definition is equivalent with saying that every countable infinite subset of X has a limit point, and I used this in my proof.

 

[facenian]

Note that that definition is equivalent with saying that every countable infinite subset of X has a limit point, and I used this in my proof.

Yes, it cast some doubt at the beginig, then I realized its ok. By the way I did try the same approach as your solution however I did not realized $S$ is closed and failed. Also not using T_1 made me think I was missing something important