# Can open sets be an open cover, for itself?

I have Baby Rudin's book with me and it clearly defines a cover to be open. In a followup, it defines a set $K$ to be compact if every open cover of $K$ contains a finite subcover.

And the rest I quote

More explicitly, the requirement is that if $\{ G_{\alpha}\}$ is an open cover of $K$, then there are finitely many indices $\alpha_1, \dots, \alpha_n$ such that $$K \subset G_{\alpha_1} \cup \dots\cup G_{\alpha_n}$$

From the notation, it would seem to suggest that we can't have $K$ to be an improper subset of the covers. Now in another book (which I referenced another user whom referenced the book) by Richardson (I don't know the full name sorry), it's definition of a cover allows a set to cover itself.

So what subtlety could I have possibly overlooked? Sorry if this mysterious reference is vague, but I couldn't get enough information on the book.

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What part of the notation suggests that $K$ can't be an improper subset? – Ben West Jan 1 '13 at 5:15
Note that a set can cover itself, but the requirement that a set be compact is that EVERY open cover has a finite subcover, not just there exists an open cover with a finite subcover. – Clayton Jan 1 '13 at 5:16
@BDub, isn't $\subset$ mean a proper subset? Mean $K$ can't be equal to the collection of open covers? – Hawk Jan 1 '13 at 5:43
@sizz Usually $\subset$ also means improper set inclusion. If one wants to be specific that a subset is a proper subset, the notation $\subsetneq$ is sometimes used. Here Rudin uses $\subset$ to include the case of improper inclusion. – Ben West Jan 1 '13 at 5:51
Some people use $\subset$ for proper inclusion and $\subseteq$ for possibly improper inclusion. Other people use $\subsetneq$ for proper inclusion and $\subset$ for possibly improper inclusion. Because of this ambiguity in the meaning of $\subset$, I don't use that symbol at all; I use only $\subsetneq$ and $\subseteq$, whose meanings are unambiguous. And of course, I recommend that everybody follow my convention. – Andreas Blass Jan 1 '13 at 20:31

$K$ can be an improper subset in the definition. An open set obviously forms an open cover of itself, and is obviously a finite cover; but a set $K$ is compact if [and only if] every open cover has a finite subcover (as you yourself wrote, as per Definition 2.32 in Baby Rudin).

You'll find soon enough (if you have not already encountered this) that the definition of compactness you discuss above is the "foundational" definition of compactness used in Rudin. You'll see that chapter 2 proceeds to show that there are ways (using Theorems introduced and proven) to establish that a set is compact if it is, for example, (i) a closed subset of a compact space, or (ii) if it is in $\mathbb{R}^n$ and is closed and bounded ... etc.

I think the main confusion here is with the notation Rudin uses to denote "is a subset of."

Usually, authors (like Richardson) use $\subseteq$ to denote subset or set inclusion in general (meaning "is a proper subset of or is equal to") and they use $\subset$ exclusively to denote "is a (strictly) proper subset of".

Other authors, like Rudin, simply use $\subset$ to denote the inclusion relation: "is a subset of or is equal to". Some, but not Rudin in PMA, use the symbol "$\subsetneq$" to denote the exclusion of set equality (i.e., to denote a proper subset). See the "List of Special Symbols" which immediately precedes the "Index" in Baby Rudin: Rudin makes no mention of, nor any use of, the symbol $\subseteq$ nor of the symbol $\subsetneq$ in his text.

So to specify "$A$ is a proper subset of $B$", Rudin will specify in some sense, that he is referring to a "proper subset" or otherwise omit the prospect of set equality, e.g. "$A\subset B$ AND $A \neq B$"). That is, unless Rudin states otherwise, assume $A\subset B$ means $A\subseteq B$ when reading the text.

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@sizz Is this making sense now? Look through Rudin... – amWhy Jan 1 '13 at 20:34
I think the main problem was me was to get used to seeing this misleading notation on subsets. As for my original question, I feel like I asked "Is an open cover an open cover?". I wonder if that is an equivalent question. – Hawk Jan 4 '13 at 3:27
Sort of equivalent ;-) – amWhy Jan 6 '13 at 0:15
I have an older edition of Rudin, in the appendix, he says his notation is the "inclusion" notation. I am guessing that means he means to include equality? – Hawk Jan 6 '13 at 4:59
Yes, sizz, Rudin uses $\subset$ in its "inclusive" sense, meaning what we normally see as $\subseteq$. I took me some getting-used-to when I first encountered Rudin's subset notation. At any rate, is this question (posted) cleared up for you? – amWhy Jan 6 '13 at 18:58

$K$ can be an improper subset in the definition. Some authors use $\subseteq$ and $\subset$ for subset and proper subset respectively, while others use $\subset$ and $\subsetneq$ respectively. Rudin uses the latter.

An open set obviously forms an open cover of itself, but a set is compact if every open cover has a finite subcover. You need to understand the notation for subset and definition of compact to avoid confusion.

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That's right, so doesn't that mean K can never be equal to the collection of open covers by the notation in Rudin? – Hawk Jan 1 '13 at 5:41
So then he is implying that a set can't cover itself no? – Hawk Jan 1 '13 at 5:46
And therefore it supposedly covers itself? Because the notation says no – Hawk Jan 1 '13 at 5:51

Note that a set can cover itself, but the requirement for a set to be compact states that every possible open cover has a finite subcover. As an example take $I=[0,1]\subseteq\mathbb{R}$ endowed with the subspace topology. Then every open cover $\{G_\alpha\}_{\alpha\in A}$ of $I$ will necessarily properly contain $I$ as $0$ and $1$ must be covered by some open intervals, say $(-\delta,\delta)$ and $(1-\varepsilon,1+\varepsilon)$ for $\delta,\varepsilon>0$. This says that $I\subsetneq \bigcup_{\alpha\in A}G_\alpha$ for every open cover, but since we're in a metric space and $[0,1]$ is closed and bounded, it is compact. Thus, there are a finite number of indices, say $\alpha_1,\ldots,\alpha_n$ such that $$I\subsetneq G_{\alpha_1}\cup\cdots\cup G_{\alpha_n}.$$As an alternative, if we consider $X=[0,1]$ as a topological space in its own right, then it is compact and hence every open cover $\{U_\alpha\}_{\alpha\in A}$ of $X$ will be such that $$X=\bigcup_{\alpha\in A}U_\alpha.$$ This shows that, depending on the space and whether or not it is viewed as a subspace, an open cover may properly contain the space or may equal the space.

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Remember that the Heine-Borel theorem (which you used to say that $I$ is compact) itself requires the compactness $I$. Of course it is always possible to use a fact to prove a theorem, then forget this fact and only remember the theorem, and then restore that fact from the theorem again :-) But I'd say that boundedness and closedness of $I$ is more of an indicator of compactness than the reason for this. – Stefan Hamcke Jan 1 '13 at 18:41
Oh, and Heine-Borel only works in $\mathbb R^n$, not in every metric space. – Stefan Hamcke Jan 1 '13 at 18:45

If $X$ is a metric space and $K$ is a subset, then I believe Rudin (in his Principles book) defines $K$ to be compact if for every collection of open sets $\{U_i:i\in I\}$ of $X$ with the property that $K\subseteq\bigcup_iU_i$, there are finitely many $i_1,\ldots,i_n\in I$ with $K\subseteq U_{i_1}\cup\cdots\cup U_{i_n}$. That is to say, the definition involves coverings by sets open in $X$. Compact subsets in metric spaces are closed, and if the metric space is connected, like $\mathbf{R}^n$, non-empty, proper closed sets are not open, so, because an arbitrary union of open sets is open, if $X$ is connected and $K$ is a non-empty, proper closed subset, you will never have $K=\bigcup_iU_i$ for $U_i$ open in $X$.

The aspect of this definition that I don't really like is that it makes it seem like compactness is a relative property, i.e., that it depends somehow on the ambient space $X$. This is not the case. By restricting the metric on $X$ to $K$, $K$ becomes a metric space in its own right, and one can check that a subset $A$ of $K$ is open in the metric topology of $K$ if and only if $A=K\cap U$ for some open subset $U$ of $X$. With this in mind, you can verify that $K$ is compact in Rudin's sense given above if and only if for every collection $\{V_j:j\in J\}$ of sets open in $K$ with the property that $K=\bigcup_jV_j$ (note the equality), there are finitely many $j_1,\ldots,j_n\in J$ with $K=V_{j_1}\cup\cdots\cup V_{j_n}$.

This carries over to arbitrary topological spaces without change (I just phrased it in terms of metric spaces since I think that is the generality of Rudin's book). The point is that compactness is an intrinsic property of topological spaces, and is in no way relative to some larger ambient space.

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