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  "body": "\n## Counterexample\n\nLet\n$$T = \\left\\{ \\left( x, \\sin \\tfrac{1}{x} \\right) : 0 \u003c x \\le 1 \\right\\} \\cup \\{ (0, 0) \\} \\subseteq \\mathbb{R}^2,$$\nwith the [[subspace topology|0D5YHQGDJY]] of the plane.\nThen $T$ is [[connected|0R37DFJKFW]] but not [[path-connected|0K0GE6BCYW]]. Thus connected spaces need not be path-connected.\n\n## Remarks\n\nThe graph part $G$ is connected, as the continuous image of $(0, 1]$. The origin is in the closure of $G$, and a set between a connected set and its closure is connected. Thus $T$ is connected.\nNo path joins the origin to a point of $G$. Near $x = 0$, the curve goes up and down between $-1$ and $1$ infinitely often, and a path cannot follow it continuously.\nSome texts use the closed version, which adds the whole segment $\\{0\\} \\times [-1, 1]$. It is compact, connected and not path-connected.\nThe space $T$ is also not locally connected.\n\n## Explanations\n\n### L1 human\n\nThe curve $\\sin(1/x)$ oscillates faster and faster as $x$ goes to $0$. The origin is a limit of the curve, so the space is in one piece.\nBut you cannot walk along the curve to the origin, because the curve never settles down.\n",
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