Mathematics III - Foundations of Analysis - Summer 24

The main result in this section the continuous image of a connected set is connect. If you google connected sets you'll find many different kinds of conectedness.  Most of them have a cleaner proof of this property.  So We try to find the idea behind the proof.

Let us look at a simpler case where we consider connected metric spaces which is not defined in the book.  However, I found that it is easier to see the main idea of the proof if we start with the connected metric space.

Definition. A metric space $X$ is called disconnected it if there exist two disjoint non-empty open sets $U$ and $V$ in $X$  such that $X=U\cup V$.

Theorem. Assume that $X$ and $Y$ are metric spaces and $X$ is connected. Then if $f: X\to Y$ is a continuous function, then $f(X)$ (as a metric space induced by the metric on $Y$) is connected.

Proof.   Without loss of generality we assume that $Y=f(X)$. If $f(X)$ were not connected then there are non-empty open sets $U,V\subset X$ with $U\cap V=\emptyset$ such that $f(X)=U\cup V$.  Then $X=f^{-1}(U)\cup f^{-1}(V)$, $f^{-1}(U)\cap f^{-1}(V)=\emptyset$.  Since $f(X)=Y$, none of  the sets $f^{-1}(U)$ and $f^{-1}(V)$ is empty.  Since $f$ is continuous, $f^{-1}(U)$ and $f^{-1}(V)$ are open.  So $X$ is a union of two disjoint non-empty open sets.  This contradicts the connectedess of $X$.  ♦

Several remarks are in order.

• In the proof we assumed that $f(X)=Y$.  This is bacause $f$ as a function from $X$ to $Y$ is continuous precisely when it is continuous as function from $X$ to $f(X)$. 
• The definition of connected metric spaces is a negative one.  The positive version (used in practice) is if we write a connected metric space as a disjoint union of two open subsets, then one of the subsets must be empty.  Write the proof of the above Theorem in terms of positive version.
• Note that if the metric space $X$ is the disjoint union of the open stes $U$ and $V$, then $U$ is complement  of $V$ which means that $U$ is closed.  On the other hand, if $U$ is both open and closed, then $X$ is the union of $U$ and its complement $V$ which is also open.  A metric space is thus connected precisely when it does not contain any proper non-empty "clopen" (closed and open) subsets.
  
• It is not necessary to use the definition of connected sets  introdued in Rudin's.  We can instead use the one given here.  To this end we need to show that these two definitions are equivalent.

Proposition.  Let $X$ be a metric space and the subset $E\subset X$.  Then $E$ is connected as metric space (with the metric induced from $X$) precisely when it is a connected set of $X$.

Proof.  Assume that $E=A\cup B$ with $A$ and $B$ separated (definition of Rudin).  Define $U'=X\setminus \bar A$ and $V'=X\setminus \bar B$. Now let $U=U'\cap E$ and $V=V'\cap E$.  Since $U',V'$ are open in $X$, $U,V$ are open in $E$.  Thus $B\subset U$ for $B$ and $\bar A$ are disjoint.  By definition of $U$ we see immediately that $U\cap A=\emptyset$.  So $U=B$ since $E$ is the disjoint union of $A$ and $B$.  Similarly, we get $V=A$. Therefore $E$ is the disjoint union of the two open non-empty subsets of $U$ and $V$ (the definition introduced above).

Conversely, assume that $E$ is the disjoint union of the two open non-empty subsets of $U$ and $V$.  We want to prove that they are separated, i.e. .  $\bar V\cap U=\emptyset$, and $\bar U\cap V=\emptyset$. 

For each $x\in U$ there is a $r_x>0$ so that $B_E(x,x_r):=\{y\in E: d(x,y)$ $\Rightarrow$ $B_E(x,x_r)\cap V=\emptyset$  $\Rightarrow$ $B_X(x,x_r)\cap V=\emptyset$ (for $B_X(x,x_r)\cap E=B_E(x,x_r)$).
Set $U'=\cup_{x\in U}B_X(x,r_x)$.  Then $U$ is an open set in $X$ and hence $\bar V\subset X\setminus U'$.  Note obviously that $U\subset U'$, so $\bar V\cap U=\emptyset$ and in the same way we can prove that $V\cap\bar U=\emptyset$.  Thus $U$ and $V$ are separated.  ♦