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I'm studying elementary topology and stumbled upon the comb space. It says in my notes that the comb space is "obviously" path connected, however I fail to see this. Of course I can create a path between every two points, but if I choose the points (0.5,0) and (0,0.5) I don't see how the function can be continuous at (0,0), which it has to pass through, as it's not a smooth transition. I imagine my mistake is understanding what continuity really is in a topological space, but can't really put my finger in it. Appreciate any help!

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    $\begingroup$ Doesn't the comb space contain the line segments from $(0,0)$ to $(1,0)$ and from $(0,0)$ to $(0,1)$? Isn't their union path-connected? Aren't your two points in this subset? $\endgroup$ – Angina Seng Oct 7 '18 at 8:15
  • $\begingroup$ It does, hence why I understand that I can create a path, I still fail to see why the function is continuous at (0,0) $\endgroup$ – Mageer Oct 7 '18 at 8:21
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    $\begingroup$ Your question suggests that you are confusing "continuous" with smooth. A path does not have to a smooth curve, since it doesn't have to be differentiable. Your intuitive sense that you can't have "sharp angles" isn't right, here. $\endgroup$ – Matt Oct 7 '18 at 8:22
  • $\begingroup$ Ahhh, my bad, I'm an idiot, yes, I somehow managed to confuse differentiability with continuity! Must be too early for math :) $\endgroup$ – Mageer Oct 7 '18 at 8:23
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If you want to define a path from $(\frac12, 0)$ to $(0, \frac12)$ you can just use parametrisations of the line segments via $(0,0)$: $p_1(t) = (1-t)(\frac12,0) + t(0,0) = (\frac12(1-t), 0)$ from $[0,1]$ to $X$, is clearly continuous, and the same holds for $p_2(t) = (1-t)(0,0) + t(0, \frac12) = (0,\frac12 t)$ from $[0,1]$ to $X$. Then we can combine these paths in the usual way:

  • $p(t) = p_1(2t) = (\frac12(1-2t), 0) = (\frac12-t, 0)$ for $0 \le t \le \frac12$.
  • $p(t) = p_2(2t-1) = (0, \frac12(2t-1))= (0,t-\frac12)$ for $\frac12 \le t \le 1$.

This is continuous by the glueing or pasting lemma: $p$ is defined on two closed sets $[0,\frac12]$ and $[\frac12,1]$ separately by a continuous formula (composition of $t \to 2t$ on $[0,\frac12]$ and $p_1$ etc.) and coincide with each other on the overlap $\{\frac12\}$, as $p(\frac12) = p_1(1) = p_2(0) = (0,0)$, which is needed to make this at all well-defined as a function. So $p$ is then in total a continuous function. There is no problem of continuity at $(0,0)$ at all, it's just the standard joining point of the two separate continuous paths in $X$.

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  • $\begingroup$ See here for a proof of the pasting lemma. It's also in Munkres. $\endgroup$ – Henno Brandsma Oct 7 '18 at 9:05

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