Connected versus locally connected spaces

the topologist’s sine curve.

the graph \(\theset{(x, \sin(1/x)) : x > 0}\) is connected (continuous image of \((0,\infty)\)), and its closure is \(X\), so \(X\) is connected.

Let \[ U=X\cap\bigl(\mathbb R\times(-1/2,1/2)\bigr), \] an open neighborhood of \(p\) in \(X\). We claim that the connected component of \(p\) in \(U\) is exactly \[ C=\{0\}\times(-1/2,1/2). \] Certainly \(C\) is connected and contains \(p\).

Now let \[ q=(x_q,\sin(1/x_q))\in U \] be any point of the oscillating graph, so \(x_q>0\). Choose \(m\) sufficiently large that \[ c=\frac1{\pi/2+2\pi m}<x_q. \] Then \(\sin(1/c)=1\), so \(U\) contains no point whose first coordinate is \(c\). Consequently \[ U_- = U\cap\{x<c\},\qquad U_+=U\cap\{x>c\} \] form a separation of \(U\): both are relatively open, and because \(U\) has no point with first coordinate \(c\), each is also the complement of the other in \(U\). We have \(p\in U_-\) and \(q\in U_+\). Hence no connected subset of \(U\) containing \(p\) can contain any graph point. This proves that the component of \(p\) in \(U\) is precisely \(C\).

Finally, \(C\) is not a neighborhood of \(p\) in \(X\). Every Euclidean neighborhood of \((0,0)\) meets the graph \(\{(x,\sin(1/x)):x>0\}\) at points with \(x>0\) and \(\sin(1/x)\) arbitrarily close to \(0\). Thus there is no connected open set \(W\) with \[ p\in W\subseteq U, \] because any such \(W\) would have to lie in the component \(C\) but \(C\) contains no neighborhood of \(p\). Therefore \(X\) is not locally connected at \(p\).

a disjoint union of two open intervals.

each point has a connected neighborhood (a small interval around it).

it is a disjoint union of two nonempty open sets.

1 gives a connected but not locally connected space; 2 gives a locally connected but not connected space (yes, such spaces exist).