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To prove by contradiction, suppose $X$ is not connected. Then $X$ can be written as a union of two nonempty disjoint open set $U$ and $V$. Then can I assume that there exists a continuous function from $X$ to $\mathbb{R}$, please?

user106035
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  • There is not necessarily an injective function from $X$ to $\mathbb{R}$; for if the cardinality of $X$ is strictly larger than that of $\Bbb{R}$, there are no such functions. –  Nov 07 '13 at 04:08
  • You don't need one-to-one. Define a continuous function on $X$ that takes on precisely the values $0$ and $1$. – Ted Shifrin Nov 07 '13 at 04:14
  • There you go. Now prove it's continuous. – Ted Shifrin Nov 07 '13 at 04:15
  • If I defind $f(U)=1$ and $f(V)=3$, is f a continuous function, please? – user106035 Nov 07 '13 at 04:16
  • @user106035 For every choice of open subset $A$ of $\mathbb{R}$, $f^{-1}(A)$ can only be one of four sets: $\emptyset, U, V$ or $X$. All of them are open by definition/assumption. As a result, $f$ is continuous (the topologist definition of continuous). – achille hui Nov 07 '13 at 04:24
  • If $f(U)={1}$ and $f(V)={3}$ for $U,V$ open in $X$, non-empty and $U\cup V=X$, then $f^{-1}(D)$ for $D$ open in $\Bbb R$ will be open, as either $1\in D,3\notin D$ therefor $f^{-1}(D)=U$, or $3\in D,1\notin D$ therefor $f^{-1}(D)=U$ or $f^{-1}(D)=\emptyset$ or $f^{-1}(D)=X$. – Carlos Eugenio Thompson Pinzón Nov 07 '13 at 04:25

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