Non-Homeomorphic Sets may be Homeomorphic to Subsets of Each Other

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Theorem

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $H_1 \subseteq S_1$ and $H_2 \subseteq S_2$.


Then it is possible for:

$(1): \quad T_1$ to be homeomorphic to $H_2$
$(2): \quad T_2$ to be homeomorphic to $H_1$

but:

$(3): \quad T_1$ and $T_2$ to not be homeomorphic.


Proof

Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.

Let $S_1 := \closedint 0 1$ be the closed unit interval.

Let $S_2 := \openint 0 1$ be the open unit interval.

Let $H_1 := \openint 0 1$ and $H_2 := \closedint {\dfrac 1 4} {\dfrac 3 4}$.


Then:

$(1): \quad \struct {S_1, \tau_d}$ is homeomorphic to $\struct {H_2, \tau_d}$ by the mapping $f: S_1 \to H_2$ defined as $\map f x = \dfrac x 2 + \dfrac 1 4$
$(2): \quad \struct {S_2, \tau_d}$ is trivially homeomorphic to $\struct {H_1, \tau_d}$


From Continuous Image of Compact Space is Compact, if $S_1$ and $S_2$ are homeomorphic then both must be compact.

From Closed Real Interval is Compact in Metric Space, $S_1$ is compact.

But from Open Real Interval is not Compact, $S_2$ is not compact.

Hence:

$(3): \quad \struct {S_1, \tau_d}$ and $\struct {S_2, \tau_d}$ are not homeomorphic.

$\blacksquare$


Sources

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