Let (X,T) be a topological space and define a relation ~ on X by x~y provided that there is a pathwise connected subset A of X such that x,y $\in A$. Prove that ~ is an equivalence relation on X.
I tried by assuming f(0)=x and f(1)=y and f'(0)=y and f'(1)= z. But I need to find a function, f* such that f*(0)=x and f*(1)=y.
But unable to do so. Can you please help in constructiong such a funtion.
If $f,f':[0,1] \to X$ are the equired paths define their composition
$$f \ast f'(t):=\begin{cases} f(2t) & t \in [0,\frac12]\\ f'(2t-1) & t \in [\frac12,1] \end{cases}$$
for $t \in [0,1]$. This is well-defined as $(f \ast f')(\frac12)$ is defined both as $f(1) = y$ and $f'(0)=y$. Both domain sets are closed, so by the pasting lemma we have that this function is continuous whenever $f$ and $f'$ are. So it's a continuous path from $x$ to $z$ as required.
For symmetry we use $f'(t)= f(1-t)$ when $f$ is a path from $x$ to $y$ to make one from $y$ to $x$, and a constant map suffices to show reflexivity.