3
$\begingroup$

If $X$ is connected and if $f:X\rightarrow R$ is non-constant and continuous, then $X$ is uncountable.

Proof. Since $f$ is non-constant there are $a,b\in X $ such that without loss of generality $f(a). Since $X$ is connected, we have that $f(X)$ is also connected. Since $f(X)$ is a connected subset of $R$, it is an interval (a non-degenerate one since it has at least 2 distinct elements). Since every non-degenerate interval is uncountable and $X$ gets mapped to an uncountable set, $X$ is uncountable.

  • I feel like my proof isn't professional at all. Why is every non-degenerate interval uncountable? Would I just apply Cantor's Daigonalization?
  • 3
    There’s nothing wrong with your argument, apart from the typo noted by @Alex. Every non-degenerate interval is homeomorphic to one of $(0,1)$, $[0,1)$, or $[0,1]$, and since the first of these is uncountable, they all are.2012-04-12

1 Answers 1

4

To see that a non-degenerate interval is uncountable, note that any open interval $(x,y)$ is in bijection with $(0,1)$ via $f:(x,y)\to (0,1)$ given by $f(t)=\frac{t-x}{y-x}$, and that $(0,1)$ is uncountable by Cantor Diagonalization hence $(x,y)$ is. Since any non-degenerate interval contains an open interval, we are done.

  • 0
    This may be a silly question, but is the variable $t$ in the function $f(t) = \frac{t-x}{y-x}$ in the interval $(x,y)$ or $(0,1)$ or $\mathbb{R}$?2015-02-10