i would like to understand easily notation of compact space,i had read that space is compact if it is closed and bounded,fr example following link says that The closed unit interval $[0,1]$ is compact.while The open interval $(0,1)$ is not compact:

but as i know closeness and term open could be considered differently,for example if we have set,then it is closed under addition for example,if sum of members f this set belongs to this set as well,but also we now from topology,that closed set is set which is complement of open set,and open set is member of topological space,so please help me to clarify what is meant exactly in compact topological space

  • $\begingroup$ Intuitively closed in topology means closed by limits of sequences as in your example when you cited closed by sums $\endgroup$ – user40276 Aug 23 '13 at 13:30
  • $\begingroup$ could you show me by example?i mean by numbers $\endgroup$ – dato datuashvili Aug 23 '13 at 13:33
  • $\begingroup$ $(0, 1)$ ,for instance, is not compact since you can pick the sequence $(1/n) \subset (0, 1)$ and it converges to $0 \notin (0, 1)$. $\endgroup$ – user40276 Aug 23 '13 at 13:37
  • $\begingroup$ and it is related to closeness right? $\endgroup$ – dato datuashvili Aug 23 '13 at 13:42
  • $\begingroup$ Just to note, you can close your space using nets and filters (a generalization of sequences) instead of sequences and in this case you get a "super " compact space, the Stone-Cech compactification. $\endgroup$ – user40276 Aug 23 '13 at 13:43

I think what you're confusing is the different uses of the word "closed". In calculus, analysis and topology, a subset $A$ of a space $X$ can be closed. Depending on the author, and the "strictness" of language involved, this usually means either of the three following:

  1. Boundary points of $A$ are included in $A$.
  2. Any sequence of points in $A$ which converges to a point in $X$ converges to a point in $A$.
  3. The complement $X\setminus A$ is open.

However, whenever you have an operation ("structure") on any set $A$, we say that $A$ is closed under this operation if applying the operation doesn't take you out of the set. For instance, the natural numbers aren't closed under subtraction, but the integers are. The integers again aren't closed under division, but the rational numbers are.

If the set $A$ is a closed subset of a space $X$ (as in the first paragraph), then it is for instance closed under the operation of closure (basically adding all the boundary points; if they're all in $A$ already, then it won't change anything), and under limits.

The interval $[0, 1]$, then is closed. The boundary points are $\{0\}$ and $\{1\}$, and they are part of that interval. If you have a converging sequence $x_i$ with $0 \leq x_i\leq 1$ for all $i$, then it cannot converge to a negative number, or to a number greater than $1$, so taking the limits keeps us inside our interval.

The interval $(0, 1)$ is not closed, since the boundary points are not part of the interval. Also, for instance the sequence $x_i = \frac{1}{i + 1}$ fulfills $x_i \in (0, 1)$, and it does converge, but the limit of the sequence is not contained in $(0, 1)$.

Now, for compactness, at least on the number line, in the plane, and so on (the Euclidian spaces, collectively known as $\Bbb R^n$), a set is compact iff it is closed and bounded. For general metric spaces, you need to change "bounded" to "totally bounded", and in general topology, the concept of "bounded" doesn't even make sense.

In topology and analysis, there are several very nice properties of finite sets of points which are better preserved in general by compact sets than countable, even though compact sets are not intrinsically limited in the number of points they can contain. For instance, if you have a continuous function from a space $X$ to the real number line, then if $X$ is compact, $f$ obtains a true maximum and minimum value. That would not neccessarily be the case if $X$ was countable instead.

  • $\begingroup$ that is what makes confusion in my head,i am new in topology and read from book that closed set is set, complement of which is open, and open space is all member of topological subspace $\endgroup$ – dato datuashvili Aug 23 '13 at 13:54

Closed in topology means that a space is closed under taking limits (although one must be careful with a general topological space, since limits of nets are needed. In your context, and in sufficiently nice spaces we only need to take limits of sequences). So in $\mathbb{R}^n$, a set is closed if for every sequence $(a_n)_{n\in\mathbb{N}}$ with values in your set such that $\lim_{n\to\infty}a_n$ exists, then $\lim_{n\to\infty}a_n$ is in your set. Compactness basically means that our set is closed and can't be big; that is, there aren't any sequences that can escape to infinity. So a set in $\mathbb{R}^n$ is compact if it is closed, and if each sequence at least has a subsequence that converges. So, for example, $[0,\infty)$ is closed in $\mathbb{R}$, but the sequence $a_n=n$ escapes to infinity, and all its subsequences do to.

Clarification of the term closed In many areas of mathematics, "closed" generally means that after doing some operation in some space, we stay in the same space. So closed under sum means that after adding two elements in a space, we stay in the same space. Closed in the sense of topology means that after taking a limit in the space, we stay in the space.


Limits of sequences only tend to do what you want in metric spaces. In a general topological space, you really do need to deal with nets or filters to get that "closed under limits" sense.

A different (but equivalent) notion of "closed" in topology is "closed under a Kuratowski closure operator" (you can look those up on Wikipedia):

The "closed and bounded" definition of compactness only works for $\Bbb R^n$. In the more general context of metric spaces, compactness can be defined as "complete and totally bounded". However, in general, compactness is usually defined thus:

A set $S$ is compact iff whenever $\mathcal F$ is a set of open sets such that $S\subseteq\bigcup\mathcal F$, there exists a finite subset $\mathcal G$ of $\mathcal F$ such that $S\subseteq\bigcup\mathcal G$.

Equivalently (given the axiom of choice for the first, and a weaker principle called the ultrafilter lemma for the others), every net in $S$ has a convergent subnet, or every filter on $S$ has a convergent refinement, or every ultrafilter on $S$ converges.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.