# Path connectedness of comb space

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!

• 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? – Angina Seng Oct 7 '18 at 8:15
• 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) – Mageer Oct 7 '18 at 8:21
• 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. – Matt Oct 7 '18 at 8:22
• Ahhh, my bad, I'm an idiot, yes, I somehow managed to confuse differentiability with continuity! Must be too early for math :) – Mageer Oct 7 '18 at 8:23

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 $$X$$.