myslsl

I can say "sin(1/x) is a continuous function on (0,1] but its graph is not path connected", which is more formal, but likely not mean anything to most of the reader. In that sense, I guess I have also lied :)

It's also false. Take any pair of points on the graph of sin(1/x) using the domain (0,1] that you just gave. Then we can write these points in the form (a,sin(1/a)), (b,sin(1/b)) such that 0 < a < b without loss of generality. The map f(t)=(t,sin(1/t)) on [a,b] is a path connecting these two points. This shows the graph of sin(1/x) on (0,1] is path connected.

This same trick will work if you apply it to the graph of ANY continuous map from a connected subset of R into R. This is what my graph example was getting at.

The "topologists sine curve" example you see in pointset topology as an example of connected but not path connected space involves taking the graph you just gave and including points from its closure as well.

Think about the closure of your sin(1/x) graph here. As you travel towards the origin along the topologists sine curve graph you get arbitrarily close to each point along the y-axis between -1 and 1 infinitely often. Why? Take a horizontal line thru any such point and look at the intersections between your horizontal line and your y=sin(1/x) curve. You can make a limit point argument from this fact that the closure of sin(1/x)'s graph is the graph of sin(1/x) unioned with the portion of the y-axis from -1 to 1 (inclusive).

Path connectedness fails because there is no path from any one of the closure points you just added to the rest of the curve (for example between the origin and the far right endpoint of the curve).

A better explanation of the details here would be in the connectedness/compactness chapter in Munkres Topology textbook it is example 7 in ch 3 sec 24 pg 157 in my copy.

However, I like to push back on the assumption that, in the context of teaching continuous function, the underlying space needs to be bounded: one of the first continuous function student would encounter is the identity function on real, which has both a infinite domain and range.

This is fine. I stated boundedness as an additional assumption one might require for pragmatic reasons. It's not mandatory. But it's easy to imagine somebody trying to be clever and pointing out that if we allow the domain or range to be unbounded we still have problems. For example you literally cannot draw the identity function in full. The identity map extends infinitely along y=x in both directions. You don't have the paper, drawing utensils or lifespan required to actually draw this.

More impressively, you can have function that is continuous, but you cannot find a connected path on it (i.e. not path connected). In plain words, if anyone told you "a function is continuous when you can draw it without lifting your pen". They have lied to you.

You are misrepresenting an analogy as a lie. Besides that, in the context where the claim is typically made, the analogy is still pretty reasonable and your example is just plain wrong.

People are talking about continuous maps on subsets of R into R with this analogy basically always (i.e., during a typical calc 1 or precalc class). The only real issue are domain requirements in such a context. You need connectedness in the domain or else you're just always forced into lifting your pen.

There are a couple other requirements you could add as well. You might also want to avoid unbounded domains since you can't physically draw an infinitely long curve. Likewise you might want to avoid open endpoints or else things like 1/x on (0,1] become a similar kind of problem. But this is all trivial to avoid by saying "on a closed and bounded interval" and the analogy is still fairly reasonable without them so long as you keep the connectedness requirement.

For why your example is just wrong in such a context, say we're only dealing with continuous maps on a connected subset of R into R. Recall the connected sets in R are just intervals. Recall the graph of a function f with domain X is the set {(x,f(x)) : x is in X}. Do you see why the graph of such a function is always path connected? Hint: Pick any pair of points on this graph. Do you see what path connects those two points?

Once you want to talk about continuous maps between more general topological spaces, things become more complicated. But that is not within the context in which this analogy is made.

Yes.

Even in the best case scenarios you still have "I was a modern nazi" as a documented thing you did that you could end up needing to explain for the rest of your life. How plausible does "Oh, but I was secretly one of the good guys!" sound to you?

Ross+Nicolas=Rossolas

Even if the argument doesn't persuade them at the time it still makes sense to point it out to them so that they are (hopefully) aware of it later.

Vigier is pretty famous for making fretless guitars but they are also pretty pricey afaik. It's not particularly hard to convert an existing guitar if you have any glass workers in your area willing to cut a mirror board.

I did roughly this way back in 2009 on a cheapo strat clone with a bolt-on neck:

1. Have a piece of mirror cut in the shape of the fretboard on the current neck.

2. Remove the frets from your old fretboard with pliers.

3. Fill the fret slots with wood filler.

4. Sand the whole thing down flat.

You can remove the fretboard entirely to swap it with the mirror board if you like, but sanding the whole thing down to the desired height seemed simpler to me at the time. You also retain vaguely useful "guide" marks from where the fret slots used to be with this approach.

Note that the height/width of your new board needs to play well with your nut/bridge height and whether or not you removed the old board. You also want a piece of mirror thick enough not to crack.

5. Epoxy the mirror board to the neck.

6. Sand off any excess epoxy and buff the sides smooth.

This approach worked okay for me at the time. I don't recall any exact materials or measurements I used since I did this over a decade ago. I mostly just winged it and tried to use common sense.

I will say the whole process is pretty finicky. A lot of small things contribute to playability in general. Choice of strings (roundwound, flatwound, different gauges), nut/bridge height, truss rod adjustments, neck shims etc. There's also the worry of cracking the glass from an overzealous truss rod adjustment and effectively breaking the whole neck (though this never actually happened to me).

The main issue I noticed playing fretless electric is that sustain is reduced. On a typical electric guitar the string vibrates between the metal fret and bridge materials (ignoring the nut). These materials are fairly hard, but on a fretless instrument the string vibrates between your much softer finger tips and the bridge. Perhaps a compressor pedal or some type of sustainer system would help?

If you pay attention to vigier recordings they tend to do really well with sustain. So their typical setup might be worth researching and trying to mimic.

For a toy DIY project to experiment with it's fairly fun, but I wouldn't expect anything game changing. Getting a nice sounding + nice to play set-up is challenging and involves a lot of nitty-gritty details.

As a side note, you could technically stop at step 4, though you'd probably want to sand things to a particular radius rather than flat. This is a common approach bass players take to convert fretted basses to fretless basses. There are many guides on how to do this online.

Disclaimers: This was something I did nearly 15 years ago as a teenager after reading quite a lot of random internet posts on it. Don't use my rambling as a source if you decide to try this. Use a real guide (there are many for fretted to fretless bass conversion guides that would apply for the first 4-ish steps for example). I am not responsible for gear you break or money you waste.

You could also just buy a slide for cheap if you're into that.

Hah no worries. Thanks for being so reasonable yourself lmao.

Fair points. The latter case is basically where my concern is.

I don't have a don't in this don't.

I think you are assuming a level of competence from people that I don't have faith people actually have. People absolutely can and do take "you cannot prove a negative" as a real logical rule in the literal negation sense. This isn't colloquialism. This is people misunderstanding what the phrase means.

I have definitely had conversations with idiots that have taken this phrase to mean that you just literally cannot logically prove negated statements. Whether folks like you get that that is not what the phrase refers to is irrelevant to why I'm pointing out the distinction.

If you subscribe to classical logic (i.e., propositonal or first order logic) this is not true. Proof by contradiction is one of the more common classical logic inference rules that lets you prove negated statements and more specifically can be used to prove nonexistence statements in the first order case. People go so far as to call the proof by contradiction rule "not-introduction" because it allows you to prove negated things.

Here's a wiki page that also disagrees and talks more specifically about this "principle": source (note the seven separate sources on various logicians/philosophers rejecting this "principle" as well).

If you're talking about some other system of logic or some particular existential claim (e.g. existence of god or something else), then I've got not clue. But this is definitely not a rule of classical logic.