Exercise 18.5

Question

Show that the subspace $(a,b)$ of ${\mathbb{R}}$ is homeomorphic with $(0,1)$ and the subspace $[a,b]$ of ${\mathbb{R}}$ is homeomorphic with $[0,1]$.

Dan Whitman · Accepted Answer

First we show that $(a,b)$ is homeomorphic to $(0,1)$.
  \begin{proof}
    First let $X = (a,b)$ and $Y = (0,1)$, and define the map $f: X 	o Y$ by
    \begin{gather*}
      f(x) = \frac{x - a}{b - a}
    \end{gather*}
    for any $x \in X$, noting that this is defined since $a < b$ so that $b - a > 0$.
    It is trivial to show that $f$ is a bijection.

Now, $f$ is a linear function that could just as well be defined as a map from ${\mathbb{R}}$ to ${\mathbb{R}}$, and clearly, this would be continuous by basic calculus.
    It then follows from Theorem~18.2 part~(d) that restricting its domain to $X$ means that it is still continuous.
    We also clearly have from basic algebra that its inverse is the function ${f}^{-1}: Y 	o X$ defined by
    \begin{gather*}
      {f}^{-1}(y) = a + y(b-a)
    \end{gather*}
    for $y \in Y$.
    As this is also linear, it too is continuous by the same argument.
    This suffices to show that $f$ is a homeomorphism.
  \end{proof}

The exact same argument shows that $[a,b]$ is homeomorphic to $[0,1]$ by simply setting $X = [a,b]$ and $Y = [0,1]$ in the above proof.
  It is assumed that here again $a < b$ even though the interval $[a,b]$ is valid if $a = b$ and simply becomes $[a,b] = [a,a] = \left\{aight\}$.
  However, clearly, this set cannot be homeomorphic to $[0,1]$ since it is finite whereas $[0,1]$ is uncountable.

Exercise 18.5

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Exercise 18.5

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