# Compact Topological Spaces .... Stromberg, Theorem 3.36 .... ....

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In summary: Updated - - -For each $k$, $U_{y_k}$ is disjoint from $U_{y_k}$. Therefore $U$ is disjoint from $V_{y_k}$ (because $U\subseteq U_{y_k}$). That holds for each $k$, and so $U$ is disjoint from \bigcup_{k=1}^n V_{y_k} = V. But $S\subseteq V$, so $U$ is disjoint from $S$, in other words $U\subseteq S'$.
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I am reading Karl R. Stromberg's book: "An Introduction to Classical Real Analysis". ... ...

I am focused on Chapter 3: Limits and Continuity ... ...

I need help in order to fully understand the proof of Theorem 3.36 on page 102 ... ... Theorem 3.36 and its proof read as follows:

View attachment 9134

In the above proof by Stromberg we read the following:

" ... ...Next let $$\displaystyle U = \bigcap_{ k = 1 }^n U_{ y_k }$$. Then $$\displaystyle U$$ is a neighbourhood of $$\displaystyle x$$ and $$\displaystyle U \subset S'$$ ... "
My question is as follows:

It seems plausible that $$\displaystyle U \subset S'$$ ... ...

... ... BUT ... ...

... how would we demonstrate rigorously that $$\displaystyle U \subset S'$$ ... ... ?

(Note that $$\displaystyle S'$$ is $$\displaystyle S$$ complement ...)

Help will be much appreciated ... ...

Peter=================================================================================The above post mentions Hausdorff spaces ... so I am providing access to Stromberg's definition of a Hausdorff space ... as follows:
View attachment 9135I believe it may be helpful to MHB readers to have access to some of Stromberg's terminology and notation associated with topological spaces ... so I am providig access to the same ... as follows:
View attachment 9136

Hope that helps ... ...

Peter

#### Attachments

• Stromberg - Theorem 3.36 ... .png
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• Stromberg - Defn 3.20 ... Defn of a Hausdorff Space .... .png
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• Stromberg - Defn 3.11 ... Terminology for Topological Spaces ... .png
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Hi Peter,

If $U\not\subset S'$, there is an element $u\in U\cap S$. Since the $V_y$ cover $S$, $u\in V_{y_n}$ for some $n$.

On the other hand, as $u\in U$, $u\in U_{y_k}$ for all $k$. In particular, $u\in U_{y_n}$ and this contradicts the fact that $U_{y_n}$ and $V_{y_n}$ are disjoint.

Note that the fact that $S$ is compact is essential (the theorem would obviously be false otherwise); do you see why?

For each $k$, $U_{y_k}$ is disjoint from $U_{y_k}$. Therefore $U$ is disjoint from $V_{y_k}$ (because $U\subseteq U_{y_k}$). That holds for each $k$, and so $U$ is disjoint from $$\displaystyle \bigcup_{k=1}^n V_{y_k} = V$$. But $S\subseteq V$, so $U$ is disjoint from $S$, in other words $U\subseteq S'$.Edit. Sorry, I didn't see that castor28 had already replied.

castor28 said:
Hi Peter,

If $U\not\subset S'$, there is an element $u\in U\cap S$. Since the $V_y$ cover $S$, $u\in V_{y_n}$ for some $n$.

On the other hand, as $u\in U$, $u\in U_{y_k}$ for all $k$. In particular, $u\in U_{y_n}$ and this contradicts the fact that $U_{y_n}$ and $V_{y_n}$ are disjoint.

Note that the fact that $S$ is compact is essential (the theorem would obviously be false otherwise); do you see why?

Thanks castor28 ... ...

You write:

" ... ... Note that the fact that $S$ is compact is essential (the theorem would obviously be false otherwise); do you see why? ... ... "Intriguing question ... Not sure I see why ... can you help ...

Mind you ... the argument given depends on having a finite subcover of S which arises via compactness ... but do we really need a finite cover ... would an infinite cover work with slightly amended arguments ... again ... I am not sure ...Peter

- - - Updated - - -

Opalg said:
For each $k$, $U_{y_k}$ is disjoint from $U_{y_k}$. Therefore $U$ is disjoint from $V_{y_k}$ (because $U\subseteq U_{y_k}$). That holds for each $k$, and so $U$ is disjoint from $$\displaystyle \bigcup_{k=1}^n V_{y_k} = V$$. But $S\subseteq V$, so $U$ is disjoint from $S$, in other words $U\subseteq S'$.Edit. Sorry, I didn't see that castor28 had already replied.

Thanks for the hep, Opalg ...

Peter

Hi Peter,

You need that to ensure that $U$ is a neighborhood of $x$. In a topological space, any union or any finite intersection of open sets is open, but that does not hold for infinite intersections.

For example, in $\mathbb{R}$, the intersection of the open intervals $\{(-x,x)\mid x\ne0\}$ is the singleton $\{0\}$, which is not open.

castor28 said:
Hi Peter,

You need that to ensure that $U$ is a neighborhood of $x$. In a topological space, any union or any finite intersection of open sets is open, but that does not hold for infinite intersections.

For example, in $\mathbb{R}$, the intersection of the open intervals $\{(-x,x)\mid x\ne0\}$ is the singleton $\{0\}$, which is not open.

Thanks for that really subtle and interesting point, castor28 ...

Peter

## 1. What is a compact topological space?

A compact topological space is a mathematical concept in topology that describes a space where every open cover has a finite subcover. In simple terms, this means that the space is "small" enough that any collection of open sets that covers it can be reduced to a finite number of open sets that still cover the space.

## 2. Can you give an example of a compact topological space?

Yes, the interval [0,1] with the standard topology is an example of a compact topological space. This means that any open cover of [0,1] can be reduced to a finite number of open sets that still cover the interval.

## 3. What is the significance of compact topological spaces?

Compact topological spaces have many important applications in mathematics, particularly in analysis and geometry. They allow for the study of continuity, convergence, and other important properties of functions and spaces. Additionally, many theorems and results in mathematics rely on the compactness of certain spaces.

## 4. What is Theorem 3.36 in Stromberg's book about compact topological spaces?

Theorem 3.36 in Stromberg's book states that in a compact topological space, every infinite subset has a limit point. This means that in a compact space, there is always a point that is "close" to an infinite subset, regardless of how the subset is constructed.

## 5. Are all compact topological spaces finite?

No, not all compact topological spaces are finite. In fact, there are many examples of infinite compact spaces, such as the Cantor set and the real line with the finite complement topology. However, it is true that all finite topological spaces are compact.

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