Proof. If \(X\) is path-connected, then for every \(y\in X\) there is a path in \(X\) from \(x\) to \(y\). Assume that for every \(y\in X\) there is a path in \(X\) from \(x\) to \(y\). Let \(y,z\in X\). We show that there is a path in \(X\) from \(y\) to \(z\). Let \(f\) be a path in \(X\) from \(x\) to \(y\) and \(g\) be a path in \(X\) from \(x\) to \(z\). Define \(h:\bra {0,1}\to X\) by

\[ h\of x:=\begin {cases} f\of {2x} & \text {if }0\le x\le \frac {1}{2};\\ g\of {2\cur {\frac {1}{2}-x}} & \text {if }\frac {1}{2}\le x\le 1. \end {cases} \]

Then \(h\) is a path in \(X\) from \(y\) to \(z\).  □

Proof. If \(y\) belongs to such a union \(Y\), then \(y\) belongs to a subspace \(Z\) of \(X\) that is path-connected and contains \(x\). Consequently, there is a path \(f\) in \(Z\) from \(x\) to \(y\). Then \(f\) is a path in \(Y\) from \(x\) to \(y\).  □

Proof. Let \(X\) be path-connected and suppose, for a contradiction, that \(\set {A,B}\) is a separation of \(X\). Let \(x\in A\) and \(y\in B\) and \(f\) be a path in \(X\) from \(x\) to \(y\). Then \(\set {f^{-1}\bof A,f^{-1}\bof B}\) is a separation of \(\bra {0,1}\), which is a contradiction since \(\bra {0,1}\) is connected.  □

Proof. Since

\[ X’:=\set {\ang {x,y}\in \br ^{2}:x\in \ocl {0,1},\,y=\sin \frac {1}{x}} \]

is the image of continuous function on a connected space \(\ocl {0,1}\), it follows that \(X’\) is connected. Since \(X\) is the closure of \(X’\) in \(\br \), it is connected.

Suppose, for a contradiction, that \(X\) is path-connected. Let \(f\) be a path in \(X\) from \(\ang {0,0}\) to \(\ang {1/\pi ,0}\). Let

\[ t:=\sup \set {s\in \bra {0,1}:\cur {f_{1}}\of s=0}, \]

where \(f_{1}\) is the composition of \(f\) with the projection on the first coordinate. We get a contradiction by showing that \(f_{2}\) is not continuous at \(t\), where\(f_{2}\) is the composition of \(f\) with the projection on the second coordinate.

Since \(f_{1}\) is continuous, it follows that \(f_{1}\of t=0\) so \(t<1\). By the intermediate value property of the function \(f_{1}\) there are sequences \(\cur {c_{n}}_{n\in \bn }\) and \(\cur {d_{n}}_{n\in \bn }\) in \(\ocl {t,1}\) both converging to \(t\) such that \(f_{2}\of {c_{n}}=1\) and \(f_{2}\of {d_{n}}=-1\) for each \(n\in \bn \). This proves the discontinuity of \(f_{2}\) and provides the required contradiction.  □

Proof. Let \(x,y\in X\) and \(g\) be a path in \(X\) from \(x\) to \(y\). Then \(f\circ g\) is a path in \(Y\) from \(f\of x\) to \(f\of y\).  □

Proof. Let \(X_{\ga }\) be path-connected for each \(\ga \in A\) and \(X:=\prod _{\ga \in A}X_{\ga }\). Let \(x:=\cur {x_{\ga }}_{\ga \in A}\) and \(y:=\cur {y_{\ga }}_{\ga \in A}\) be point in \(X\). For each \(\ga \in A\) there is a path \(f_{\ga }\) in \(X_{\ga }\) from \(x_{\ga }\) to \(y_{\ga }\). Then

\[ f:=\prod _{\ga \in A}f_{\ga }, \]

defined by

\[ f\of t:=\cur {f_{\ga }\of t}_{\ga \in A}, \]

for each \(t\in \bra {0,1}\), is a path in \(X\) from \(x\) to \(y\).  □

Sketch of proof. Let \(X_{n}:=\set {0,1}\) with the discrete topology and \(X:=\prod _{n\in \bn }X_{n}\). There is a continuous surjection \(g:X\to \bra {0,1}\) defined by

\[ g\of {\cur {x_{n}}_{n\in \bn }}:=\sum _{n\in \bn }\frac {x_{n}}{2^{n}}. \]

Then \(g\tm g:X^{2}\to \bra {0,1}^{2}\) is a continuous surjection. There is a homeomorphism \(h:X\to X^{2}\) defined by

\[ h\of {\cur {x_{n}}_{n\in \bn }}:=\ang {\cur {x_{2n}}_{n\in \bn },\cur {x_{2n+1}}_{n\in \bn }}. \]

and there is a homeomorphism \(\ph :X\to C\), where \(C\sub \bra {0,1}\) is the Cantor set, defined by

\[ \ph \of {\cur {x_{n}}_{n\in \bn }}=\sum _{n\in \bn }\frac {2x_{n}}{3^{n}}. \]

Then \(\ph ^{-1}:C\to X\) can be extended to a continuous function \(\psi :\bra {0,1}\to X\) by defining \(\psi \) to be linear on each open interval removed during the construction of \(C\). Then

\[ f:=\cur {g\tm g}\circ h\circ \psi \]

is a continuous surjection \(\bra {0,1}\to \bra {0,1}^{2}\) as required.  □