I am attempting to prove the Intermediate Value Theorem from the proof of clopen sets in $\mathbb{R}$. I have several lemmas and a corollary inbetween which do most of the heavy work (I've excluded their proofs to save space).

Have I missed anything, and or are the assumptions/proofs valid?

$\textbf{Proposition.}$ The only clopen subsets of $\mathbb{R}$ are the $\emptyset$ and $\mathbb{R}$ itself.

$\textit{Proof.}$ Let $Y$ be the subset $Y \subset \mathbb{R}$ and $\chi_{Y}$ a characteristic function $\chi_{Y} : \mathbb{R} \to \{0, 1\}, y \mapsto \begin{cases} 1 & y \in Y \\ 0 & y \notin Y \end{cases} $.

Let $A \subset \mathbb{R}$ be an open subset (or subsets). Then the characteristic function $\chi_{Y}$ is continuous iff the preimage $I^{-1} \equiv \chi_{y}^{-1}(A)$ is open. If elements $0,~1 \in A$ then $I^{-1} = \emptyset$, and is open. If $0,~1 \notin A$ then $I^{-1} = \mathbb{R}$, which is also open. If $0 \notin A$ and $1 \in A$ then $I^{-1} = Y$, and must be open. Similarly, If $0 \in A$ and $1 \notin A$ then $I^{-1} = \mathbb{R} \setminus Y$, and must also be open. But then $Y$ must be closed by definition. Therefore, the only clopen subsets of $\mathbb{R}$ are the $\emptyset$ and the set $\mathbb{R}$ itself.

$\textbf{Corollary (Connectedness).} ~\textit{The set $W$ is connected iff there are subsets $X \subset W$ and $Y \subset W$} \\ \textit{such that $X$ and $Y$ are clopen.}$

Now for IVT...

$\textbf{Lemma 1.}~\textit{Let $a$ and $b$ be any elements in $\mathbb{R}$. Intervals of form} \\ \textit{$[a,b]$, $(a,b]$, $[a,b)$, $(a,b)$ are the only subsets in $\mathbb{R}$ that are connected.}$

$\textbf{Lemma 2.}~\textit{If a connected space is continuous then so is its image.}$

$\textbf{Intermediate Value Theorem.} ~\textit{If $f : \mathbb{R} \to \mathbb{R}$ is a function which is continuous at every point}$ $\textit{of the interval $[a,b]$ and $f(a) < 0, f(b) < 0$ then $f(x) = 0$ at some point $x \in (a,b)$.}$

$\textit{Proof.}$ By Lemma 1, the interval $[a,b]$ is connected. Then by Lemma 2, the image $f([a,b])$ is also connected. Therefore, there exists some point $x \in (a,b)$.

  • $\begingroup$ Your attempted proof that $Y\subset \mathbb R$ is clopen iff $Y\in \{ \emptyset ,\mathbb R \}$ only shows that if $Y$ is clopen and if $\chi_Y$ is continuous then $Y\in \{\emptyset, \mathbb R\}.$ But what if $\chi_Y$ is not continuous? $\endgroup$ – DanielWainfleet Dec 13 '16 at 14:54
  • $\begingroup$ I also see that your attempted proof that $Y\subset \mathbb R$ is clopen iff $Y\in \{\emptyset, \mathbb R\} $ is circular, as it contains the sentence "Thus $Y$ is a clopen set, which is possible only if $Y$ is to only $\emptyset$ or $\mathbb R.$" . So you have stated that what you are trying to prove is true, in order to prove it. $\endgroup$ – DanielWainfleet Dec 13 '16 at 15:08
  • $\begingroup$ @user254665 I've made some edits and reduced repetition. Things should be more clear. I don't see the issue with the characteristic function not being continuous, care to elaborate? If it is not continuous then you can't show that $Y \in \{\emptyset, \mathbb{R}\}$. However if it is continuous there are 2 clopen subsets in $\mathbb{R}$. Which is what I needed to prove. $\endgroup$ – user334916 Dec 14 '16 at 3:58
  • 1
    $\begingroup$ Up to the last sentence of that proof, you have shown that $\chi_Y$ is continuous iff $Y$ is clopen. Which is also valid if you replace $\mathbb R$ with any discrete space $X,$ in which all subsets are clopen. The last sentence does not follow from the previous ones. $\endgroup$ – DanielWainfleet Dec 14 '16 at 12:14

In the corollary are you saying that the only connected subset of $\mathbb{R}$ are $\mathbb{R}$ and $\emptyset$? But in lemma 1 you say $[a,b]$ is connected.

Intervals of the form $(a,b), [a,b), (a,b]$ are also connected.

  • $\begingroup$ Thanks for pointing that out. Made a mistake. It should read: The $\mathbb{R}$ are connected iff there is some subset $W \subset \mathbb{R}$ such that $W$ is clopen, $W = \emptyset$ or $W = \mathbb{R}$ itself. $\endgroup$ – user334916 Dec 12 '16 at 3:48
  • $\begingroup$ To the OP: I can't follow your comment. I think you have some typos in it. $\endgroup$ – DanielWainfleet Dec 13 '16 at 14:57

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