I am confusing about sphere minus points. [I mean $S^2 \setminus \{ points \}$]

First, , I understand $S^2$ $\setminus$ {North pole} homeomorphic $\mathbb{R}^2$. [I can understand this using sterographic projection]

Next, $S^2$ $\setminus$ {North pole, South pole} is a homeomorphic to Cylinder ($S^1 \times \mathbb{R}$), actually geometrically(I mean by drawing) I can understand this procedure. Of course I can explicitly find the homeomorphism. [for example, intuitively Why do we have $S^2-\{N,S\}\simeq S^1\times\mathbb{R}$? or constructing homeomorphism, How can I prove that a cylinder is diffeomorphic to a twice-punctured $n$-sphere? ]

Now I want further, As a next step I tried to understand three points case.

I edit question in a following way.

Since $S^2 \setminus$ {three points} is homotpoic to figure 8 (not homeomoprhic). I am concern with its homeomorphic shape.

Is there any name for $S^2 \setminus$ {three points}? I simply guess it is homeomorphic to $\mathbb{R}^2 \setminus$ {two points}, but can not draw its shape in my head.

and how about its generalization?

$S^2 \setminus $ {more than three points} $\simeq ? $

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    $\begingroup$ A sphere minus $3$ points is not homeomorphic to a figure $8$, but it is homotopy equivalent to that space. $\endgroup$ – Cheerful Parsnip Jun 12 at 15:40
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    $\begingroup$ The sphere minus three points is homotopy equivalent to a figure eight. By the way, it doesn't matter which point you remove, or which two points. $S^2\setminus \{x_1,x_2,\dots,x_n\}$ choosing $n$ distinct points will return the same space, up to homomorphism, for any two ways to pick the $x_i.$ $\endgroup$ – Thomas Andrews Jun 12 at 15:46
  • $\begingroup$ @CheerfulParsnip, Thomas Andrews, I see, I might mistype the note. $\endgroup$ – phy_math Jun 12 at 15:47
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    $\begingroup$ The sphere minus $n$ points, for $n>1$ is homotopy equivalent to $$\underbrace{S_1\vee S_1\vee\cdots\vee S_1}_{n-1\text{ times}}$$ $\endgroup$ – Thomas Andrews Jun 12 at 15:52
  • $\begingroup$ @ThomasAndrews, Oh that's a good information! Thanks! $\endgroup$ – phy_math Jun 12 at 15:54

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