Completely Regular Space

A topological space $$X$$ is said to be a completely regular space if every closed set $$A$$ in $$X$$ and a point $$x \in X$$, $$x \notin A$$, then there exists a continuous function $$f:X \to \left[ {0,1} \right]$$, such that $$f\left( x \right) = 0$$ and $$f\left( A \right) = \left\{ 1 \right\}$$.

In other words, a topological space $$X$$ is said to be a completely regular space if for any $$x \in X$$ and a closed set $$C$$ not containing $$x$$, there exists a continuous function $$f:X \to \left[ {0,1} \right]$$ such that $$f\left( x \right) = 0$$ and $$f\left( C \right) = 1$$.

Remark:

Let us consider a continuous function $$g:X \to \left[ {0,1} \right]$$ defined as $$g\left( x \right) = 1$$ and $$g\left( A \right) = 0$$. Since the constant function is continuous therefore taking the function $$g\left( x \right) = I\left( x \right) – f\left( x \right)$$, where $$I\left( x \right) = 1,{\text{ }}\forall x \in X$$.

Now

\[ g\left( x \right) = I\left( x \right) – f\left( x \right) = 1 – 0 = 1 \]

And
\[ g\left( A \right) = I\left( A \right) – f\left( A \right) = 1 – 1 = 0 \]

Moreover the continuous function defined in the condition for a completely regular space is said to separate point $$x$$ from the set $$A$$.

 

Tychonoff Space

A completely regular $${T_1}$$ space is said to be a Tychonoff space or a $${T_{3\frac{1}{2}}}$$-space.

Note: It may be noted that since the product of $${T_1}$$ space is a $${T_1}$$ space and the product of a completely regular space is a completely regular space, so the product of a Tychonoff space is a Tychonoff space.

 

Theorems
• Every completely regular space is a regular space as well.
• Every completely regular $${T_1}$$ space is a Hausdorff space or $${T_2}$$ space.
• Every subspace of a completely regular space is a completely regular space.
• The product of a completely regular space is a completely regular space.
• Every subspace of a Tychonoff space is a Tychonoff space.
• Every Tychonoff space is a Hausdorff space.