Let $X=C([0,1])$ be the space of all continuous real valued function on $[0,1]$. Equip $X$ with the $\sup$ metric, $d(f,g)=\sup_{x\in[0,1]}|f(t)-g(t)|$
Let,
$S=\{f\in X| f(0)\neq f(1)\}$
$T=\{f\in X|f(0)\leq f(1)\}$
Are $S$ and $T$ connected?
This is a completely odd problem for me. So far I only came across problems where $X$ was some subset of $\mathbb{R}^n$. How to deal with function spaces?
Thanks for reading.
On
The set $S$ is disconnected. Consider the map$$\begin{array}{rccc}\varphi\colon&S&\longrightarrow&\mathbb R\\&f&\mapsto&f(1)-f(0).\end{array}$$Then the range of $f$ is $\mathbb{R}\setminus\{0\}$, which is disconnected. Since $\varphi$ is continuous, this proves that $S$ is disconnected.
But $T$ is connected. It is actually path-connected: if $f,g\in T$, the range of the path$$\begin{array}{rccc}\gamma\colon&[0,1]&\longrightarrow&\mathcal{C}[0,1]\\&\lambda&\mapsto&\lambda f+(1-\lambda)g\end{array}$$is a subset of $T$. Besides, $\gamma(0)=g$ and $\gamma(1)=f$.
In this kind of situation, remember the definition of a connected set : A topological space $X$ is said to be disconnected if it is the disjoint union of two open sets (its equivalent to be the disjoint union of two closed sets, and it is also equivalent to have a continuous surjective application from X to $\{0,1\}$) ; it is said to be connected otherwise.
It is not obvious to understand why such a definition encapsulates the intuitive idea of connexity. The basic idea is the following lemma, linked to the definition : if $X$ is connex included in $Y$, if $X$ meets a set $C$ and its complementary, it meets its boundary. So you cannot use a set $C$ to "separate" $X$ in two parts. It would be possible to show this : $X$ is connex if and only if for all embedding of $X$ in a topological space $Y$, for all set $C$ included in $Y$, the preceding property is valid. Intuitively this is clearer, but the first definition is simpler in the sense you do not have to check all the embeddings $X$ in $Y$ and all the subsets $C$ of $Y$...
With this definition, it is quite easy, as we will see, to show that a set is disconnected.
To show the connectedness, a nice way is to show the path-connectedness. It is a more intuitive notion and easier to prove. A topological set $X$ is path-connected if for all $x,y \in X$ there exists $\gamma$ from $[0,1]$ to $X$ continuous such that $\gamma(0) = x, \gamma(1) = y$ (you trace a path in $X$ linking $x$ and $y$). you can show that path-connectedness imply connectedness.
You can also define the notion of connex component of a $x\in X$ : it is the biggest connex set containing $x$ (this is well define because a union of connex sets which intersects non trivially is connex ; hence, the connex component of $x$ is simply the union of all the connex sets containing $x$). Quite useful if you want to be more precise.
Remember also some properties (you can try to show these by using definitions) : - The direct image of a connected set by a continuous application if a connected set. - More generally, the direct image of a topological X set by a continuous application has less connex components than X. - The reciprocal image of a space $Y$ by a continuous application has more connex components than $Y$.
It is quite powerful to prove results about connexity : you just need to exhibit some continous application to have information.
For the first example, there is several ways to reason. - You can remark that S_+ = { f(1) > f(0) } and $S_- = \{ f(1) < f(0) \}$ are open and disjoint ; so you're done. - You can use the continuous application $f \rightarrow f(1) - f(0)$ as in the José Carlos Santos' post.
You can even show that $S_+$ and $S_-$ are (path-)connex.
In the second example, you also prove that it is (path-)connex as in Carlos Santos' post.