Is the space C[0,1] connected?

For the proof I followed similar steps to theorem 45.7:

Suppose C[0,1] is not connected. Then there exists open nonempty subsets U and V of C[0,1] s.t $C[0,1] = U\cup V$ and $U\cap V = \emptyset$. Let $x=(x_1,...,x_n)\in U$ and $y=(y_1,...,y_n)\in V$. Let $$f(t) = (tx_1+(1-t)y_1,tx_2+(1-t)y_2,...,tx_n+(1-t)y_n)$$ for $0\leq t \leq1$ . We show that f is continuous from [0,1] into C.

Let $\epsilon >0$. Let $\delta = \epsilon /(\sup_{x \in [0,1]}|f(x)-g(x)|)$. If $|t_1 - t_2| < \delta$, then

$d(f(t_1),f(t_2)) < \epsilon$. Then f is continuous.

But I am not sure if the steps are true or this theorem can be applied for C[0,1].

  • 1
    $\begingroup$ What does $x=(x_1,...,x_n)\in U$ mean? Elements of $U$ are functions, not $n$-tuples... $\endgroup$ – Eric Wofsey Nov 20 '17 at 1:12

Let $f$ be a continuous function defined on $[0,1]$, $f(x)=tf(x)$, $t\in [0,1]$ defines a path between $f$ and the $0$ function, so $C([0,1])$ is pathe connected hence connected.


For $f,g\in C[0,1]$, for any fixed $t\in[0,1]$, then $tf+(1-t)g\in C[0,1]$, so $C[0,1]$ is convex and hence connected.


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