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One way to create a function with a hole is multiplying and dividing it by x, like this:

$f(x) = 1$

$g(x) = \frac{x}{x}$

This creates a so-called "removable singularity" at $x=0$. At school I was taught not to remove such singularities, so I wonder if holes are ever real, or are they just a mathematical artefact that has nothing to do with reality?

Other types of singularities make perfect sense (for example in the trigonometric function $tan$), but they don't create holes. I'm not familiar with any situation where a hole in the function makes sense.

potato
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    Can you clarify your definition of "real function" or "function that has to do with reality"? – angryavian Jan 04 '20 at 16:27
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    "reality" does not belong to the scope of mathematical reasoning. –  Jan 04 '20 at 16:30
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    I think you should first try to understand what a "function" is. – Qi Zhu Jan 04 '20 at 16:30
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    Why does $\tan$ not "create holes" ? –  Jan 04 '20 at 16:31
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    @YvesDaoust it creates an asymptote, not a hole that you can plug a number into it to get a continuous function. – potato Jan 04 '20 at 16:32
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    Do you think that an asymptote "has to do with reality" ? –  Jan 04 '20 at 16:36
  • I mean that the presence of a hole is somehow meaningfully describing the thing that this function was written to describe. For example $tan$ function's asymptote describes the fact that you can't have a triangle with two 90 degree angles. Can a hole tell something about the thing it describes or can it always be ignored? – potato Jan 04 '20 at 16:38
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    @potato, well $x\mapsto x/x$ having "a hole" describes the fact that you can't divide by $0$. – Ennar Jan 04 '20 at 16:42
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    This may help clarify things. Why do you think the type of singularities in the function $\tan x,$ or $1/x$ make perfect sense? What about $\sin x/ x,$ which is also a hole like the one in $x/x$? These are the things you need to clarify before one may even begin to approach this question more seriously. – Allawonder Jan 04 '20 at 16:45
  • @Ennar I wrote 1 ↦ x/x, and the inability to divide by zero is what creates this hole, but that doesn't mean the limit of the function at x=0 doesn't describe the thing this function was written to describe. – potato Jan 04 '20 at 16:45
  • @potato, I have no idea what you are talking about at all. – Ennar Jan 04 '20 at 16:50
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    As I already mentioned, my best suggestion would be for you to learn the definition of a function because neither $f$ nor $g$ that you mentioned are functions. What is the domain? What is the codomain? – Qi Zhu Jan 04 '20 at 18:18

1 Answers1

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A function can be described as a set $F$ of ordered pairs with the property that if $(x,y)\in F$ and $(x,z)\in F$, then $y=z$. The idea is that it is the collection of $\{(x,f(x)\mid x\in D\}$ where $D$ is the domain of the function.

A "hole" in a real input, real-valued function (a subset of $\mathbb R\times \mathbb R$) amounts to the absence of a particular ordered pair.

For example:

$F_1=\{(x,\frac{x}{x})\mid x\in \mathbb R, x\neq 0\}$

$F_2=\{(x,1)\mid x\in \mathbb R\}$

are almost, but not quite, the exact same function. The difference is that $F_2$ has one more point: $(0,1)$, whereas $F_1$ has a hole there.

rschwieb
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