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math.colostate.edu/~clayton/courses/500/500_4.pdf
(a) Show that no two of the spaces $(0,1),(0,1]$, and $[0,1]$ are homeomorphic.
Proof. Suppose $(0,1)$ and $(0,1]$ are homeomorphic, with the homeomorphism given by $f$. If $A=(0,1)-\left\{f^{-1}(1)\right\}$, then $\left.f\right|_A: A \rightarrow(0,1)$ is a homeomorphism, by Theorem 18.2(d). However, the interval $(0,1)$ is connected, whereas
\[
A=\left(0, f^{-1}(1)\right) \cup\left(f^{-1}(1), 1\right)
\]
is not connected, so $A$ and $(0,1)$ cannot be homeomorphic. From this contradiction, then, we conclude that $(0,1)$ and $(0,1]$ are not homeomorphic.
Similarly, suppose $g:(0,1] \rightarrow[0,1]$ is a homeomorphism. Then, if $B=(0,1]-\left\{g^{-1}(0), g^{-1}(1)\right\}$, $\left.g\right|_B: B \rightarrow(0,1)$ is a homeomorphism. However, $g^{-1}(0) \neq g^{-1}(1)$, so at most one of these can be 1 , meaning one must lie in the interval $(0,1)$. Suppose, without loss of generality, that $g^{-1}(0) \in(0,1)$. Then
\[
B=\left(0, g^{-1}(0)\right) \cup\left(g^{-1}(0), 1\right]-\left\{g^{-1}(1)\right\}
\]
is not connected, whereas $(0,1)$ is, so the two cannot be homeomorphic. From this contradiction, then, we conclude that $(0,1]$ and $[0,1]$ are not homeomorphic.
A similar argument easily demonstrates that $(0,1)$ and $[0,1]$ are not homeomorphic, so we see that no two of the spaces $(0,1),(0,1]$, and $[0,1]$ are homeomorphic.
(b) Suppose that there exist imbeddings $f: X \rightarrow Y$ and $g: Y \rightarrow X$.
Show by means of an example that $X$ and $Y$ need not be homeomorphic.
Example: Let $f:(0,1) \rightarrow[0,1]$ be the canonical imbedding and let $g:[0,1] \rightarrow(0,1)$ such that
\[
g(x)=\frac{x}{3}+\frac{1}{3}
\]
Then $g([0,1])=\left[\frac{1}{3}, \frac{2}{3}\right] .$ Obtain $g^{\prime}$ by restricting the range of $g$ to $g([0,1])=$ $\left[\frac{1}{3}, \frac{2}{3}\right]$. We claim that $g^{\prime}$ is a homeomorphism. Since multiplication and addition are continuous, as are the inclusion map and compositions of continuous functions, we see that $g^{\prime}$ is continuous, as is $g^{\prime-1}$, where
\[
g^{\prime-1}(x)=3\left(x-\frac{1}{3}\right)
\]
Both of these maps are also bijective, so we see that $g^{\prime}$ is indeed a homeomorphism, meaning $g$ is an imbedding.
However, as we saw in part (a) above, $(0,1)$ and $[0,1]$ are not homeomorphic, so two spaces need not be homeomorphic for each to be imbedded in the other.
(c) Show $\mathbb{R}^n$ and $\mathbb{R}$ are not homeomorphic if $n>1$.
Lemma 0.1. If $f: X \rightarrow Y$ is a homeomorphism and $X$ is path-connected, then $Y$ is path-connected.
Proof. Let $x, y \in Y$. Then there exists a continuous path $g:[a, b] \rightarrow X$ such that $g(a)=f^{-1}(x)$ and $g(b)=f^{-1}(y)$. Define
\[
h:=f \circ g .
\]
Then $h:[a, b] \rightarrow Y$ is continuous and
\[
h(a)=(f \circ g)(a)=f(g(a))=f\left(f^{-1}(x)\right)=x
\]
and
\[
h(b)=(f \circ g)(b)=f(g(b))=f\left(f^{-1}(y)\right)=y
\]
since $f$ is bijective. Hence, $h$ is a path from $x$ to $y$, so, since our choice of $x$ and $y$ was arbitrary, $Y$ is path connected.
Proposition 0.2. $\mathbb{R}^n$ and $\mathbb{R}$ are not homeomorphic if $n>1$.
Proof. Suppose $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is a homeomorphism. Then, restricting the domain to $\mathbb{R}^n-\{0\}$ gives a homeomorphism of the punctured euclidean space to $\mathbb{R}-\{f(0)\}$. However, the punctured euclidean space is path-connected (as shown in Example 4), whereas $\mathbb{R}-\{f(0)\}$ is not even connected, let alone path-connected. To see this, we need only note that
\[
\mathbb{R}-\{f(0)\}=(-\infty, f(0)) \cup(f(0), \infty)
\]
so the open sets $(-\infty, f(0))$ and $(f(0), \infty)$ give a separation of this space. Hence, by the above lemma, the punctured euclidean space and $\mathbb{R}-\{f(0)\}$ are not homeomorphic, a contradiction. Therefore, we conclude that $\mathbb{R}^n$ and $\mathbb{R}$ are not homeomorphic. |
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