Connected and functionally Hausdorff with at least two points implies cardinality at least that of the continuum: Difference between revisions

Revision as of 00:02, 28 January 2012

This article gives the statement, and possibly proof, of a topological space satisfying certain conditions (usually, a combination of separation and connectedness conditions) is uncountable.

Statement

Any connected functionally Hausdorff space having at least two points is uncountable. In fact, its cardinality is at least equal to the cardinality of the continuum.

Definitions used

Connected space

Further information: Connected space

Functionally Huasdorff space

Further information: functionally Hausdorff space

A topological space $X$ is termed a functionally Huasdorff space if, given any two points $x,y\in X$ , there is a continuous function $f:X\to [0,1]$ such that $f(x)=0$ and $f(y)=1$ .

Related facts

Connected normal implies uncountable: This follows because, by Urysohn's lemma, normal spaces are Urysohn.
Connected regular Hausdorff implies uncountable: Interestingly, this proof does not yield that the cardinality must be at least that of the continuum, only that it must be uncountable.

Proof

Suppose $X$ is a connected functionally Hausdorff space with at least two points. Say, $x\neq y\in X$ are two points. Then, by the functionally Hausdorff condition, there exists a function $f:X\to [0,1]$ such that $f(x)=0$ and $f(y)=1$ .

Now, we claim that $f$ is surjective. Suppose not; suppose there exists $a\in (0,1)$ such that $f^{-1}(a)$ is empty. Then, $f^{-1}[0,a)$ and $f^{-1}(a,1]$ are disjoint open subsets whose union is $X$ , and both are nonempty (because $f(x)=0$ and $f(y)=1$ . This contradicts the assumption that $X$ is connected, hence $f$ must be surjective.

Thus, the cardinality of $X$ must be at least that of the continuum.

@@ Line 3: / Line 3: @@
 ==Statement==
-Any [[connected space|connected]] [[Urysohn space]] having at least two points is uncountable. In fact, its cardinality is at least equal to the [[cardinality of the continuum]].
+Any [[connected space|connected]] [[functionally Hausdorff space]] having at least two points is uncountable. In fact, its cardinality is at least equal to the [[cardinality of the continuum]].
 ==Definitions used==
@@ Line 11: / Line 11: @@
 {{further|[[Connected space]]}}
-===Urysohn space===
+===Functionally Huasdorff space===
-{{further|[[Urysohn space]]}}
+{{further|[[functionally Hausdorff space]]}}
-A [[topological space]] <math>X</math> is termed a Urysohn space if, given any two points <math>x,y \in X</math>, there is a continuous function <math>f:X \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(y) = 1</math>.
+A [[topological space]] <math>X</math> is termed a functionally Huasdorff space if, given any two points <math>x,y \in X</math>, there is a continuous function <math>f:X \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(y) = 1</math>.
 ==Related facts==
 * [[Connected normal implies uncountable]]: This follows because, by [[Urysohn's lemma]], normal spaces are Urysohn.
-* [[Connected regular implies uncountable]]: Interestingly, this proof does not yield that the cardinality must be at least that of the continuum, only that it must be uncountable.
+* [[Connected regular Hausdorff implies uncountable]]: Interestingly, this proof does not yield that the cardinality must be at least that of the continuum, only that it must be uncountable.
 ==Proof==
-Suppose <math>X</math> is a connected Urysohn space with at least two points. Say, <math>x \ne y \in X</math> are two points. Then, by the Urysohnness, there exists a function <math>f:X \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(y) = 1</math>.
+Suppose <math>X</math> is a connected functionally Hausdorff space with at least two points. Say, <math>x \ne y \in X</math> are two points. Then, by the functionally Hausdorff condition, there exists a function <math>f:X \to [0,1]</math> such that <math>f(x) = 0</math> and <math>f(y) = 1</math>.
 Now, we claim that <math>f</math> is surjective. Suppose not; suppose there exists <math>a \in (0,1)</matH> such that <math>f^{-1}(a)</matH> is empty. Then, <math>f^{-1}[0,a)</math> and <math>f^{-1}(a,1]</math> are disjoint open subsets whose union is <math>X</math>, and both are nonempty (because <math>f(x) = 0</math> and <math>f(y) = 1</math>. This contradicts the assumption that <math>X</matH> is connected, hence <math>f</math> must be surjective.
 Thus, the cardinality of <math>X</math> must be at least that of the continuum.