Compact Sets and Closed Sets in R^n .... .... D&K Lemma 1.8.2 .... ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of the proof of Lemma 1.8.2 ... ...

Duistermaat and Kolk"s Lemma 1.8.2 and the preceding definition and notes on compactness read as follows:View attachment 7720Just prior to the statement of Lemma 1.8.2 we read (referring to Lemma 1.8.2):

" ... ... It is immediate from Definition 1.8.1 and Lemma 1.2.12 ... ... "Lemma 1.2.12 reads as follows:View attachment 7721Now ... Definition 1.81 gives as a characterization of a compact set \(\displaystyle K\) that every sequence of elements in \(\displaystyle K\) contains a subsequence which converges to a point in \(\displaystyle K\) ... ...

while

Lemma 1.2.12 (iii) gives as a characterization of a set K that is closed in \(\displaystyle K\) that every sequence \(\displaystyle ( x_k)_{ k \in \mathbb{N} }\) of points \(\displaystyle x_k \in K\) that is convergent to a limit, say \(\displaystyle a \in \mathbb{R}^n\), we have \(\displaystyle a \in K\) ...Now as I see it ... for Lemma 1.8.2 to be true these statements must mean the same thing (or at least imply each other ...) ... but they do not seem to mean exactly the same thing ...Can someone resolve and clarify this issue ...

Help will be appreciated ...

Peter

 

[GJA]

Hi, Peter.

but they do not seem to mean exactly the same thing ...

Your intuition here is correct. The Lemma's statement assumes the sequence you start with is convergent. The Definition's statement starts with an arbitrary sequence (i.e., it may or may not converge).

For example, consider the set $I=[0,\infty)$. Then the sequence $x_{n}=n$ is a sequence in $I$ that admits no convergent subsequences (any subsequence will be unbounded and thus divergent). The set $I$ is closed in the sense of the Lemma however.

 

[Math Amateur]

Hi, Peter.
Your intuition here is correct. The Lemma's statement assumes the sequence you start with is convergent. The Definition's statement starts with an arbitrary sequence (i.e., it may or may not converge).

For example, consider the set $I=[0,\infty)$. Then the sequence $x_{n}=n$ is a sequence in $I$ that admits no convergent subsequences (any subsequence will be unbounded and thus divergent). The set $I$ is closed in the sense of the Lemma however.

Thanks for the help GJA ... ... appreciate the help ...

... but ... still struggling a bit ...

Would certainly help if you could indicate the main steps of the proof of Lemma 1.8.2 ... ...Hope you can help further ...

Peter

 

[GJA]

Hi, Peter. I accidentally missed the first question in your post asking for help with the proof.

Suppose $C$ is a compact subset of $K$. To prove that $C$ is closed using Lemma 1.2.12, let $\{x_{n}\}$ be a sequence in $C$ that converges to a point $a$. To prove $C$ is closed, we must show $a\in C$. Since $\{x_{n}\}$ is a sequence in the compact set $C$, it contains a subsequence $\{x_{n_{k}}\}$ that converges to a point $b\in C.$ Since subsequences of convergent sequences must converge to the same limit as the original sequence, we have $a=b\in C$, so that $C$ is closed.

Now suppose $C$ is a closed subset of the compact set $K$, and let $\{x_{n}\}$ be any sequence in $C$. To prove that $C$ is compact, we must show that $\{x_{n}\}$ admits a convergent subsequence whose limit is in $C$. To do so, use the fact that $C\subseteq K$ and that $K$ is compact to conclude that $\{x_{n}\}$ admits a convergent subsequence whose limit, say $a$, is in $K$ (Note: this is not the end of the argument because we need to show that $a\in C$).

Now, forget about the sequence $\{x_{n}\}$ and only think about the sequence $\{x_{n_{k}}\}$ (i.e., no need to think of it as a subsequence of anything anymore - call it $y_{k}$ if you'd like). Then the sequence $\{x_{n_{k}}\}$ is a convergent sequence in $C$. Since $C$ is closed (assumption), the limit $a\in C$.

 

[Math Amateur]

Hi, Peter. I accidentally missed the first question in your post asking for help with the proof.

Suppose $C$ is a compact subset of $K$. To prove that $C$ is closed using Lemma 1.2.12, let $\{x_{n}\}$ be a sequence in $C$ that converges to a point $a$. To prove $C$ is closed, we must show $a\in C$. Since $\{x_{n}\}$ is a sequence in the compact set $C$, it contains a subsequence $\{x_{n_{k}}\}$ that converges to a point $b\in C.$ Since subsequences of convergent sequences must converge to the same limit as the original sequence, we have $a=b\in C$, so that $C$ is closed.

Now suppose $C$ is a closed subset of the compact set $K$, and let $\{x_{n}\}$ be any sequence in $C$. To prove that $C$ is compact, we must show that $\{x_{n}\}$ admits a convergent subsequence whose limit is in $C$. To do so, use the fact that $C\subseteq K$ and that $K$ is compact to conclude that $\{x_{n}\}$ admits a convergent subsequence whose limit, say $a$, is in $K$ (Note: this is not the end of the argument because we need to show that $a\in C$).

Now, forget about the sequence $\{x_{n}\}$ and only think about the sequence $\{x_{n_{k}}\}$ (i.e., no need to think of it as a subsequence of anything anymore - call it $y_{k}$ if you'd like). Then the sequence $\{x_{n_{k}}\}$ is a convergent sequence in $C$. Since $C$ is closed (assumption), the limit $a\in C$.

Thanks for the help GJA ... appreciate it ... !

Just a couple of minor questions ...

\(\displaystyle C\) is actually given as a subset of a compact set ... can we assume that a subset of a compact set is compact? ... if so ... why ... ?In Lemma 1.2.12 we are dealing with a set \(\displaystyle F\) that is closed in \(\displaystyle \mathbb{R}^n\) ... but in Lemma 1.8.2 we are dealing with a set \(\displaystyle C\) being closed ... not in \(\displaystyle \mathbb{R}^n\) ... but in a set \(\displaystyle K\) ... does this alter the proof in any way?Hope you can help clarify ...

Peter

 

[GJA]

\(\displaystyle C\) is actually given as a subset of a compact set ... can we assume that a subset of a compact set is compact? ... if so ... why ... ?

In general subsets of compact sets are not compact; e.g., $(0,1)\subset [0,1].$ However, the proof of sufficiency (i.e., the "forward direction") of Lemma 1.2.12 assumes $C$ is a compact subset of $K$.

In Lemma 1.2.12 we are dealing with a set \(\displaystyle F\) that is closed in \(\displaystyle \mathbb{R}^n\) ... but in Lemma 1.8.2 we are dealing with a set \(\displaystyle C\) being closed ... not in \(\displaystyle \mathbb{R}^n\) ... but in a set \(\displaystyle K\) ... does this alter the proof in any way?

No, the proof stays the same. It is true (from a general topological point of view) that a closed subset of a closed set is closed in the larger space.

 

[Math Amateur]

In general subsets of compact sets are not compact; e.g., $(0,1)\subset [0,1].$ However, the proof of sufficiency (i.e., the "forward direction") of Lemma 1.2.12 assumes $C$ is a compact subset of $K$.
No, the proof stays the same. It is true (from a general topological point of view) that a closed subset of a closed set is closed in the larger space.

Thanks GJA ... for all your help on this Lemma ...

Peter