Gnome Is The Only DE That Plays Nice With Zoom Out Of The Box As Of Early 2020. Source: I Tried A Lot

Gnome is the only DE that plays nice with zoom out of the box as of early 2020. Source: I tried a lot of other things first and the fucking sound didn’t work. Admittedly, every once in a while, gnome tells me that my only speaker is named Dummy Audio and when that happens I have to run pulseaudio and ^C before it finishes telling me that pulseaudio is already running. Promptly after killing the audio daemon in this way, everything works in a lovely fashion for the next day or so.

Stop talking shit about Linux Desktop Environments!

KDE is Pretty! XFCE is Efficient! Gnome... Mate is Traditional!

I think there are uncountably many homeomorphism types of countable punctured planes.

Curves in the plane with dense punctures (for example, what’s left of the unit circle after you puncture the rational points on the unit circle out of existence) are nowhere path connected in a way that non-convergent Cauchy sequences can pick up, so homeomorphisms ought to send densely punctured curves to other densely punctured curves. Then number and nesting patterns of densely punctured curves would give a nice invariant.

But in particular, after cutting out rational points from circles of integer radius about the origin, we create a bunch of un-punctured rings, only one of which has as its interior an open disk (that’s the middle one).

then simply by puncturing the rational points from a_1 disjoint loops in the first circle, a_2 from the second, a_3 from the third and so on, we can inject the integer sequence (a_n) into the set of homeomorphism classes of punctured planes. (Since countable unions of countable sets of punctures are countable, the resulting construction actually lives in the set of countably punctured planes).

That jeans post has got me thinking, is there a nice way of distinguishing between planes with infinite numbers of holes in them. More formally, can we classify up to homeomorphism infinitely punctured planes easily?

I have a feeling cardinality matters. I'd conjecture that ℝ²\(ℚ×{0}) is not homeomorphic to ℝ²\(C×{0}), where C is the middle third Cantor set. But how could we prove that?

Also I have a feel cardinality isn't the only thing that matters. I reckon density might matter too. For example, is ℝ²\(ℚ×{0}) homeomorphic to ℝ²\(ℕ×{0}). I guess the difference there is that each hole in the second space is isolated whereas those in the first aren't.

It'd be interesting to hear what other people think :))

Hi, I'm

Math enjoyers do not DNI. If you like math you are required to interact with me.

mods do a sudo rm -rf /home/her

mods compress her into a .tar.gz

I'm in this picture and I don't like it