Mathematical Analysis

Consider \(a=\inf S\in [-\infty,\infty)\text{,}\)\(b=\sup S\in (-\infty,\infty]\text{.}\) It suffices to show \((a,b)\subseteq S\text{,}\) since then \(S\subseteq [a,b]\) and \(S\cap((-\infty,a)\cup(b,\infty))=\emptyset\text{.}\) To this end, suppose \(\exists \, t\in (a,b),t\notin S\text{.}\) Then \(S\subseteq (-\infty,t)\cup(t,\infty)\text{.}\) Both \((-\infty,t)\) and \((t,\infty)\) are open sets. Since \(t>a=\inf S,\exists\, s\in S\) with \(s<t\text{.}\) Then \(s\in S\cap(-\infty,t),\) so \(S\cap(-\infty,t)\) nonempty. Similarly, \(S\cap (t,\infty)\neq \emptyset\text{.}\) This shows \(S\) is disconnected, a contradiction. Hence \((a,b)\subseteq S\text{.}\)

Suppose ftsoc that \([0,1]\subseteq \mathbb{R}\) is disconnected. Then there exist open sets \(U,V\subseteq [0,1]\) such that \([0,1]=U\cup V, U\cap V=\emptyset, U\neq\emptyset\) and \(V\neq \emptyset\text{.}\) Wlog, \(0\in U\) (has to be in one of them). Let \(C=\{{t\in[0,1]}\,:\,{[0,t]\subseteq U}\}\text{.}\) It's enough to show \(1\in C\text{,}\) since then \(U=[0,1]\) and \(V=\emptyset\text{,}\) a contradiction. Note that \(0\in C\) so \(C\neq \emptyset\text{.}\) Let \(t_0=\sup C\in [0,1]\text{.}\)

Claim 1: \(t_0\in C\text{.}\) Suppose not. Then \([0,t_0]\not\subseteq U\text{.}\) As \(t_0=\sup C, \exists\, (t_n)_{n=1}^\infty\subseteq C\) so that \(t_1<t_2<\dots\) and \(\lim_{n\rightarrow\infty} t_n=t_0\text{.}\) Then \([0,t_n]\subseteq U\) for all \(n\geq 1\text{,}\) and so \([0,t_0)=\cup_{n=1}^\infty[0,t_n]\subseteq U\text{.}\) This implies \(t_0\notin U\) (since \([0,t_0]\not\subseteq U\)). So \(t_0\in V\text{.}\) Since \(t_0\in V,\,\exists\,\delta>0\) with \((t_0-\delta,t_0+\delta\cap [0,1]\subseteq V\text{.}\) Let \(t_1=\max\{{t_0-\frac{\delta}{2},\frac{t_0}{2}}\}\text{.}\) Note that \(t_0>0\) since \(0\in U,t_0\notin U\text{.}\) Then \(t_1\in(t_0-\delta,t_0)\cap[0,1]\text{.}\) Also, \([0,t_1]\subseteq [0,t_0)\subseteq U\text{.}\) So \(t_1\in U\cap V\text{,}\) and hence \(U\cap V\neq \emptyset\text{.}\)

Claim 2: \(t_0=1\text{.}\) Suppose \(t_0<1\text{.}\) As \(t_0\in C,\exists \varepsilon>0\) such that \((t_0-\varepsilon,t_0+e)\cap [0,1]\subseteq U\text{.}\) Let \(t_2=\min\{{t_0+\frac{\varepsilon}{2},\frac{1+t_0}{2}}\}\text{.}\) Then \(t_2>t_0\) and \(t_2\in(t_0-\varepsilon,t_0+\varepsilon)\cap [0,1]\subseteq U\text{.}\) Note that \([t_0,t_2]\subseteq (t_0-\varepsilon,t_0+\varepsilon)\cap [0,1]\subseteq U\) and as \(t_0\in C,[0,t_0]\subseteq U\text{.}\) So \([0,t_2]=[0,t_0]\cup [t_0,t_2]\subseteq U\text{.}\) Now, \(t_2\in C\) and \(t_2>t_0=\sup C\text{,}\) a contradiction.

This shows \(1\in C\text{.}\) Hence \(U=[0,1],V=\emptyset,\) a contradiction

Suppose \(Y\) is disconnected. There are open sets \(U,V\subseteq Y\) such that \(U\cup V=Y,U\cap V=\emptyset, U\neq \emptyset, V\neq \emptyset\text{.}\) Then \(f^{-1}(U),f^{-1}(V)\subseteq X\) are open as \(f\) is continuous. Then \(f^{-1}(U)\cup f^{-1}(V)=f^{-1}(U\cup V)=f^{-1}(Y)=X\) and \(f^{-1}(U)\cap f^{-1}(V)=f^{-1}(U\cap V)=f^{-1}(\emptyset)=\emptyset\text{.}\) To show \(X\) is disconnected, we need to show \(f^{-1}(U)\neq \emptyset\) and \(f^{-1}(V)\neq \emptyset\text{.}\) As \(U\) is nonempty, exists \(y\in U\text{.}\) Since \(f\) is surjective, exists \(x\in X\) with \(f(x)=y\in U\text{.}\) Then \(x\in f^{-1}(U)\text{.}\) This shows \(f^{-1}(U)\neq \emptyset\text{.}\) Similarly, \(f^{-1}(V)\neq \emptyset\text{.}\)

WLOG let \(f(a)\leq f(b)\text{.}\) As \(f\) is continuous and \([a,b]\) is connected, \(f([a,b])\) is connected by the previous theorem. So \(f([a,b])\) is an interval. So \([f(a),f(b)]\subseteq f([a,b])\text{.}\) So if \(t\in [f(a),f(b)],\) there is \(c\in [a,b]\) with \(f(c)=t\text{.}\)

(\(\implies\)) Suppose \(S\) is disconnected in \(X\text{.}\) So exist open \(U,V\subseteq X\) with \(S\subseteq U\cup V, U\cap V=\emptyset,U\cup S\neq \emptyset,V\cap S\neq \emptyset\text{.}\) Then \(U_0:=U\cap S, V_0:=V\cap S\) are open subsets of \(S\) (open in \(S\)). Also, \(S=U_0\cup V_0,U_0\cap V_0=\emptyset,U_0\neq \emptyset, V_0\neq \emptyset\text{,}\) so \(S\) is disconnected as a metric space.

(\(\impliedby\)) Suppose \(S\) is disconnected as a metric space. There are open sets \(U_0,V_0\subseteq S\) with \(S=U_0\cup V_0, U_0\cap V_0=\emptyset,U_0\neq \emptyset, V_0\neq \emptyset\text{.}\)

Exercise: Exist open sets \(U',V'\subseteq X\) with \(U_0=S\cap U'\text{,}\) \(V_0=S\cap V'\text{.}\)

Then \(S\subseteq U'\cup V', U'\cap V'\cap S= \emptyset, U'\cap S\neq\emptyset,V'\cap S\neq\emptyset\text{.}\) (If \(U'\cap V'=\emptyset\text{,}\) we're done.) Since \(U'\cap V'\cap S=\emptyset,\) we have that \(U'\cap S\subseteq X\setminus V'\text{.}\) As \(V'\) is open, \(X\setminus V'\) is closed, and so \(\overline{U'\cap S}\subseteq X\setminus V'\text{.}\) Therefore, \((\overline{U'\cap S})\cap V'=\emptyset\text{.}\) In particular, \((\overline{U'\cap S})\cap (V'\cap S)=\emptyset\text{.}\) Similarly,\((U'\cap S)\cap (\overline{V'\cap S})=\emptyset\text{.}\) Apply the key lemma with \(A=U'\cap S,B=V'\cap S\) to get open sets \(U,V\subseteq X\) with \(U'\cap S\subseteq U, V'\cap S\subseteq V\text{,}\) and \(U\cap V=\emptyset\text{.}\) Then \(S\subseteq (U'\cup V')\cap S=(U'\cap S)\cup (V'\cap S)\subseteq U\cup V\text{.}\) Also, since \(\emptyset\neq U'\cap S\subseteq U\cap S\text{,}\) we have \(U\cap S\neq \emptyset\text{.}\) Similarly \(V\cap S\neq \emptyset\text{.}\) So \(S\) is disconnected in \(X\text{.}\) ◻