Connected and functionally Hausdorff with at least two points implies cardinality at least that of the continuum

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property must also satisfy the second topological space property
View all topological space property implications | View all topological space property non-implications
|

Property "Page" (as page type) with input value "{{{stronger}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.Property "Page" (as page type) with input value "{{{weaker}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.

Statement

Any connected Urysohn space having at least two points, is uncountable. In fact, its cardinality is at least that of the continuum.

Definitions used

Connected space

Further information: Connected space

Urysohn space

Further information: Urysohn space

A topological space $X$ is termed a Urysohn 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 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 Urysohn space with at least two points. Say, $x\neq y\in X$ are two points. Then, by the Urysohnness, 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.