# Limit point compact

In mathematics, a topological space $X$ is said to be limit point compact 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 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 $\mathbb {R}$ of all real numbers with its usual topology, since the integers are an infinite set but do not have a limit point in $\mathbb {R}$; (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 T$_1$ 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 $X=\mathbb {Z} \times Y$ where $\mathbb {Z}$ 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. This space is not T$_0$. It is limit point compact because every nonempty subset has a limit point.
* An example of T$_0$ space that is limit point compact and not countably compact is $X=\mathbb {R} ,$ the set of all real numbers, with the right order topology, i.e., the topology generated by all intervals $(x,\infty ).$ The space is limit point compact because given any point $a\in X,$ every $x<a$ 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 $X=\mathbb {Z} \times Y$ with $\mathbb {Z}$ discrete and $Y$ indiscrete as in the example above, the map $f=\pi _{\mathbb {Z} }$ given by projection onto the first coordinate is continuous, but $f(X)=\mathbb {Z}$ is not limit point compact.
* A limit point compact space need not be pseudocompact. An example is given by the same $X=\mathbb {Z} \times Y$ with $Y$ indiscrete two-point space and the map $f=\pi _{\mathbb {Z} },$ whose image is not bounded in $\mathbb {R} .$
* 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. 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 $(X,\tau ).$