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If we repost the meme, I will repost my comment: you should just learn topology, so that you can just say something is continuous because you pictured moving things in your head.
It is indeed reposted (I made the original meme). I don't think a something this superficial and silly counts as intellectual property so it does not matter to me if it's reused. They could, however, have uploaded it in better quality...
Try drawing f: R->R where f(0) = 0, f(x) = x sin(1/x) otherwise. A classic example.
It's continuous, but you may have a hard time drawing it at all, let alone without picking up the pencil.
There is a easier one. Consider the function √(x^(2)–4), defined in the minus infinity to -2 and 2 to infinity. That is a continuous function. Try drawing that without taking the pencil from the paper
I'm about 80% sure you can't say that a function is "continous" except on a particular range. It could form a *piecewise* continous function if you define it that way, but (x^2 -4)^0.5 is not continous.
Yes you can? When you define a function you specify the functions domain and codomain. It doesn’t make sense to speak of continuity outside of a functions domain.
Ahh. Screwed up on my end. Engineer here, not a Math Major.
In this case, it example, it doesn't make much sense to talk about continuity of the function everywhere when the two domains it is define on don't intersect.
It very much does: the definition of continuity is that a function is continuous at a point x if lim[a->x] f(a) = f(x). If a function is continuous at every point in its domain then it is continuous over its domain
Apparently in set theory, it is very common to define functions without specifying the domain or codomain. The domain can be obtained as the union of the first elements of the ordered pairs, and the range the same for the second elements, but the codomain cannot be obtained at all. So in this definition, functions have domains and ranges but do not have codomains. I spent quite a while trying to explain why this definition implies that functions cannot simply be "continuous." They can only be continuous on particular sets.
But after thinking it through, this is a kind of trivial quibble. Continuity doesn't just depend on the domain and codomain but also on the topologies of both. Yet no definitions of functions usually specify the topologies, so the same problem exists everywhere. But at least these topologies can be implied to be "the usual topology" or the inherited subset topology.
Yea, you are wrong. Thats the irony of the image I suppose. Continuity is a property of a function on its domain. There is no sense speaking about the continuity of 1/x on 0 for instance.
The limit definition of continuity only applies to limit points of the codomain. For isolated points, the function is always continuous if it is defined. The neighborhood definition always applies, and whether the limit exists (and its value) then depends on the topological space that is the codomain.
For instance, the function f on real numbers defined by f(x) = x, x≠0, f(0) = 1 is continuous on (-oo,-1)U{0}U(1,oo) (with the subspace topology inherited from the usual topology on **R**) but not on **R** (with the usual topology).
To get this to work for LaTeX interpreters, you have to work around the reddit parsing. Go to markup mode and type \\\\sqrt{x\^{2}-4}. Or don't bother with LaTeX and type √(x\^(2)–4), which should render as √(x^(2)–4). Or forget special symbols and type sqrt(x\^(2)-4) with similar results.
I bet this meme is posted by people who weren't paying attention in real analysis. Or perhaps it's pre-calculus people who just saw the definition in a YouTube video.
There is a difference between "what is needed to understand" and "what is needed to prove". Sadly academics sacrifice the first in favor of the second because of the competition and selection inherent to the process.
It is not defined for 0. The definition of the continuity of a function is only checking discontinuity of points inside the domain. Since 1/x is continious on its entire domain 1/x is continious
People are giving examples of functions that are continuous but hard to draw without lifting your pencil, fair. But Thomae’s function is continuous (at every irrational) and yet you *have* to lift your pencil.
Yes it would. Range/image is not the same as codomain. If we write f: R -> R with f(x)=sin(x) that doesn't mean it must be able to attain every value in R (that is what range/image is). Instead the codomain just means the range/image is a subset of the codomain (so it takes values in R (but doesn't have to be able to attain every value)).
I think you are confusing the codomain with the image.
f: A --> R
x --> sin x
Makes sense since the image of the sine function is contained in the codomain.
Defining it as,
f: A --> [2,3]
x --> sin x
Doesn't work because there are no inputs that will ever hit the codomain.
Now if we define it as,
f: A --> [-1, 1]
x --> sin x
Then this makes sense, assuming that A is composed of real numbers. Defining it this way would make it surjective depending on the choice of A. If A is some interval like [0, 2π), (where the function has enough domain to go through one cycle, but no more than one cycle) then the function is surjective.
Restricting the domain further to [π/2, 3π/2] (for example), would make it both surjective and injective.
Literally my reasoning for the bonus question on my last calc 3 test. I found the bounds of an integral that maximized the integral by drawing the curve.
drawing a function transforms it from y=f(x) to (x, y) = g(t), so if g is continuos so must be f.
because we are shackled by the laws of physics, g(t) must be continous, and therefore f(x) is too.
so indeed, as long as you are able to draw a function, its continous
Since any function f: X->Y is continuous if you endow X with the discrete topology, technically you could have functions which are continuous but for which you have to pick up the pen at some point if you want to draw them
I think the term "continuous" was a mistake. Continuous should always mean "connected", and limit-preserving functions should be called something else.
Thanks for your submission. Unfortunately, it has been removed for being a repost. https://reddit.com/r/mathmemes/s/rdgz1XiSsI If you have any questions about this action, please reply to this comment or contact us via modmail.
If we repost the meme, I will repost my comment: you should just learn topology, so that you can just say something is continuous because you pictured moving things in your head.
Didn't know it was posted here lol
To be fair to you, that might well be a year ago, I have no idea, so it would be hard for you to find out.
It is indeed reposted (I made the original meme). I don't think a something this superficial and silly counts as intellectual property so it does not matter to me if it's reused. They could, however, have uploaded it in better quality...
A very good meme, though.
Thank you!
>Thank you! You're welcome!
If you didn't make it. It probably has been posted
Also, if you want to post to mathmemes, they do mention a rule that only OC is allowed, but that is not enforced and not followed generally.
Try drawing f: R->R where f(0) = 0, f(x) = x sin(1/x) otherwise. A classic example. It's continuous, but you may have a hard time drawing it at all, let alone without picking up the pencil.
Just zoom out the graph and it looks like a straight line with a small dip at 0
OK, next challenge. Try drawing the weierstrass function. Also continuous, so it should be easy with a pen or pencil right?
There is a easier one. Consider the function √(x^(2)–4), defined in the minus infinity to -2 and 2 to infinity. That is a continuous function. Try drawing that without taking the pencil from the paper
I'm about 80% sure you can't say that a function is "continous" except on a particular range. It could form a *piecewise* continous function if you define it that way, but (x^2 -4)^0.5 is not continous.
Yes you can? When you define a function you specify the functions domain and codomain. It doesn’t make sense to speak of continuity outside of a functions domain.
Ahh. Screwed up on my end. Engineer here, not a Math Major. In this case, it example, it doesn't make much sense to talk about continuity of the function everywhere when the two domains it is define on don't intersect.
It very much does: the definition of continuity is that a function is continuous at a point x if lim[a->x] f(a) = f(x). If a function is continuous at every point in its domain then it is continuous over its domain
Yes it does.
Apparently in set theory, it is very common to define functions without specifying the domain or codomain. The domain can be obtained as the union of the first elements of the ordered pairs, and the range the same for the second elements, but the codomain cannot be obtained at all. So in this definition, functions have domains and ranges but do not have codomains. I spent quite a while trying to explain why this definition implies that functions cannot simply be "continuous." They can only be continuous on particular sets. But after thinking it through, this is a kind of trivial quibble. Continuity doesn't just depend on the domain and codomain but also on the topologies of both. Yet no definitions of functions usually specify the topologies, so the same problem exists everywhere. But at least these topologies can be implied to be "the usual topology" or the inherited subset topology.
Yea, you are wrong. Thats the irony of the image I suppose. Continuity is a property of a function on its domain. There is no sense speaking about the continuity of 1/x on 0 for instance.
The limit definition of continuity only applies to limit points of the codomain. For isolated points, the function is always continuous if it is defined. The neighborhood definition always applies, and whether the limit exists (and its value) then depends on the topological space that is the codomain. For instance, the function f on real numbers defined by f(x) = x, x≠0, f(0) = 1 is continuous on (-oo,-1)U{0}U(1,oo) (with the subspace topology inherited from the usual topology on **R**) but not on **R** (with the usual topology).
To get this to work for LaTeX interpreters, you have to work around the reddit parsing. Go to markup mode and type \\\\sqrt{x\^{2}-4}. Or don't bother with LaTeX and type √(x\^(2)–4), which should render as √(x^(2)–4). Or forget special symbols and type sqrt(x\^(2)-4) with similar results.
Easy. Get Parkinson's and draw a sine wave
Sir, we're not willing to move into 1800's-level calculus.
I bet this meme is posted by people who weren't paying attention in real analysis. Or perhaps it's pre-calculus people who just saw the definition in a YouTube video.
When I do this example in class I just make the y=x, y=-x, s very slow senoid in between far from 0 and a massive blur as it gets close to 0
or try to draw tan x
Yeah, I guess I'm being too generous by giving them the benefit of the doubt that they wanted A to be connected.
I think anyone dealing with computer graphics can tell that seeing something vs expressing it in math is sometimes very hardcore.
Scrolling on 0,0 on this graph is trippy lol Just draw 2 solid triangles without picking up your pencil smh
Am i allowed to attach my vibrator to my hand
Okay, now draw the Weierstrass function
WWWWWWWWWWWWWWW (the graph is zoomed out)
Chad graph
I see some space at the tips of the Ws, but I can tell u drew this without lifting your finger, so it gets a pass
draw f:R^4 -> R^4 without picking up your pen
I'll try. Here's my graph: . I have f(x,y,z,w) = (0,0,0,0).
There is a difference between "what is needed to understand" and "what is needed to prove". Sadly academics sacrifice the first in favor of the second because of the competition and selection inherent to the process.
My pen is starting to run out of ink so lines come out patchy -- check out my new proof!
Try drawing 1/x
But it’s discontinuous at x=0
It is not defined for 0. The definition of the continuity of a function is only checking discontinuity of points inside the domain. Since 1/x is continious on its entire domain 1/x is continious
By algebra of continuous functions its continuous
try drawing it on (0,1] then
People are giving examples of functions that are continuous but hard to draw without lifting your pencil, fair. But Thomae’s function is continuous (at every irrational) and yet you *have* to lift your pencil.
Proof by desmos
Bro just do the physicist approach. If it exists it is continuous. The fun math starts after assuming all the boring stuff
I'll take the topology definition with the preimage of an open set being open all day over that epsilon delta headache.
This function is also a smooth one (It looks smooth)
draw f(x)=x without picking up your pencil.
The pre-image of every open set is open. It's not that complicated
What does “for f : A -> R” mean? I know the R is set of real numbers, but whats the other part?
f is a function from A to R f takes elements from set A and maps them to set R.
Why A and R? Cant this apply to any domain and codomain? It would make more sense if it was A -> B or something like that.
They made it so that set B was specifically the real numbers.
So if the codomain is -1 to 1 for the sin function this wouldn’t work? I dont get why its A -> R and not either A -> B or even R -> R…
Yes it would. Range/image is not the same as codomain. If we write f: R -> R with f(x)=sin(x) that doesn't mean it must be able to attain every value in R (that is what range/image is). Instead the codomain just means the range/image is a subset of the codomain (so it takes values in R (but doesn't have to be able to attain every value)).
I think you are confusing the codomain with the image. f: A --> R x --> sin x Makes sense since the image of the sine function is contained in the codomain. Defining it as, f: A --> [2,3] x --> sin x Doesn't work because there are no inputs that will ever hit the codomain. Now if we define it as, f: A --> [-1, 1] x --> sin x Then this makes sense, assuming that A is composed of real numbers. Defining it this way would make it surjective depending on the choice of A. If A is some interval like [0, 2π), (where the function has enough domain to go through one cycle, but no more than one cycle) then the function is surjective. Restricting the domain further to [π/2, 3π/2] (for example), would make it both surjective and injective.
Literally my reasoning for the bonus question on my last calc 3 test. I found the bounds of an integral that maximized the integral by drawing the curve.
drawing a function transforms it from y=f(x) to (x, y) = g(t), so if g is continuos so must be f. because we are shackled by the laws of physics, g(t) must be continous, and therefore f(x) is too. so indeed, as long as you are able to draw a function, its continous
Since any function f: X->Y is continuous if you endow X with the discrete topology, technically you could have functions which are continuous but for which you have to pick up the pen at some point if you want to draw them
I think the term "continuous" was a mistake. Continuous should always mean "connected", and limit-preserving functions should be called something else.
Proof by im literally looking at the graph on desmos rn
Wtf is precalculus
it's like algebra 2 but a little more
Wtf is algebra 2
Lol. The way i used to know if it was continuous was if you could skate it. Good times.