Limit point compact

In mathematics, a topological space $X$ is said to be limit point compactThe terminology "limit point compact" appears in a topology textbook by James Munkres where he says that historically such spaces had been called just "compact" and what we now call compact spaces were called "bicompact". There was then a shift in terminology with bicompact spaces being called just "compact" and no generally accepted name for the first concept, some calling it "Fréchet compactness", others the "Bolzano-Weierstrass property". He says he invented the term "limit point compact" to have something at least descriptive of the property. Munkres, p. 178-179.Steen & Seebach, p. 19 or weakly countably compact if every infinite subset of $X$ has a limit point in $X.$ This property generalizes a property of compact spaces. In a metric space, limit point compactness, compactness, and sequential compactness are all equivalent. For general topological spaces, however, these three notions of compactness are not equivalent.

Properties and examples

In a topological space, subsets without limit point are exactly those that are closed and discrete in the subspace topology. So a space is limit point compact if and only if all its closed discrete subsets are finite.
A space $X$ is {{em|not}} limit point compact if and only if it has an infinite closed discrete subspace. Since any subset of a closed discrete subset of $X$ is itself closed in $X$ and discrete, this is equivalent to require that $X$ has a countably infinite closed discrete subspace.
Some examples of spaces that are not limit point compact: (1) The set $\Reals$ of all real numbers with its usual topology, since the integers are an infinite set but do not have a limit point in $\Reals$ ; (2) an infinite set with the discrete topology; (3) the countable complement topology on an uncountable set.
Every countably compact space (and hence every compact space) is limit point compact.
For T₁ spaces, limit point compactness is equivalent to countable compactness.
An example of limit point compact space that is not countably compact is obtained by "doubling the integers", namely, taking the product $X = \Z \times Y$ where $\Z$ is the set of all integers with the discrete topology and $Y = \{0,1\}$ has the indiscrete topology. The space $X$ is homeomorphic to the odd-even topology.Steen & Seebach, Example 6 This space is not T₀. It is limit point compact because every nonempty subset has a limit point.
An example of T₀ space that is limit point compact and not countably compact is $X = \Reals,$ the set of all real numbers, with the right order topology, i.e., the topology generated by all intervals $(x, \infty).$ Steen & Seebach, Example 50 The space is limit point compact because given any point $a \in X,$ every $x is a limit point of \{a\}.$
For metrizable spaces, compactness, countable compactness, limit point compactness, and sequential compactness are all equivalent.
Closed subspaces of a limit point compact space are limit point compact.
The continuous image of a limit point compact space need not be limit point compact. For example, if $X = \Z \times Y$ with $\Z$ discrete and $Y$ indiscrete as in the example above, the map $f = \pi_{\Z}$ given by projection onto the first coordinate is continuous, but $f(X) = \Z$ is not limit point compact.
A limit point compact space need not be pseudocompact. An example is given by the same $X = \Z \times Y$ with $Y$ indiscrete two-point space and the map $f = \pi_{\Z},$ whose image is not bounded in $\Reals.$
A pseudocompact space need not be limit point compact. An example is given by an uncountable set with the cocountable topology.
Every normal pseudocompact space is limit point compact.Steen & Seebach, p. 20. What they call "normal" is T₄ in wikipedia's terminology, but it's essentially the same proof as here.
Proof: Suppose $X$ is a normal space that is not limit point compact. There exists a countably infinite closed discrete subset $A = \{x_1, x_2, x_3, \ldots\}$ of $X.$ By the Tietze extension theorem the continuous function $f$ on $A$ defined by $f(x_n) = n$ can be extended to an (unbounded) real-valued continuous function on all of $X.$ So $X$ is not pseudocompact.
Limit point compact spaces have countable extent.
If $(X, \tau)$ and $(X, \sigma)$ are topological spaces with $\sigma$ finer than $\tau$ and $(X, \sigma)$ is limit point compact, then so is $(X, \tau).$

Notes

References

{{Munkres Topology|edition=2}}
{{cite book|last1=Steen|first1=Lynn Arthur|author-link1=Lynn Arthur Steen|last2=Seebach|first2=J. Arthur|author-link2=J. Arthur Seebach Jr.|title=Counterexamples in topology|publisher=Dover Publications|publication-place=New York|date=1995|orig-date=First published 1978 by Springer-Verlag, New York|isbn=0-486-68735-X|oclc=32311847}}
{{PlanetMath attribution|id=1234|title=Weakly countably compact}}

Category:Properties of topological spaces

Category:Compactness (mathematics)

Limit point compact

Properties and examples

See also

Notes

References