Tychonoff Space is Preserved under Homeomorphism
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Theorem
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a Tychonoff (completely regular) space, then so is $T_B$.
Proof
We have that $\struct {S, \tau}$ is a Tychonoff space if and only if:
- $\struct {S, \tau}$ is a $T_{3 \frac 1 2}$ space
- $\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space.
From $T_{3 \frac 1 2}$ Space is Preserved under Homeomorphism:
- If $T_A$ is a $T_{3 \frac 1 2}$ space, then so is $T_B$.
From $T_0$ (Kolmogorov) Space is Preserved under Homeomorphism:
- If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$.
Hence the result.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces