Is a function defined at a single point continuous?

Question

For example $f:\{0\}\to\{0\}$ defined by $f(x)=\sqrt{x}+\sqrt{-x}$ is a sum of two continuous functions and is therefore continuous, however for $f$ to be continuous, it has to have the limit $\lim\limits_{x_0\to x}f(x_0)=f(x)$ but it dosen't seem to have that limit.

This questions needs a bit more context. Are you asking as in a calculus question or as in a topology question? — Git Gud, Jan 07 '15 at 23:53
The answer to your question is yes, for the reasons explained below. However, it is important to draw attention to a difference between the way limits and continuity are taught at lower and higher levels. Most typically at lower levels, students are taught that the existence and value of a limit $\lim_{x \to a} f(x)$ depends only on values of $f(x)$ for $x \ne a$. This particular idea of a limit reflects what would be written as $\lim_{x \to a, x \ne a} f(x)$ at higher levels (in topology), as a special case of a more general concept $\lim_{x \to a, x \in A} f(x)$. The entire collection of... — user204305, Jan 08 '15 at 05:48
...definitions concerning limits and continuity given at lower levels is generally not conceived in order to make sense for anything other than a function whose domain is an interval, perhaps minus a finite number of points, whereas the definitions given in topology are much more general. It is not surprising then that you are encountering difficulty in understanding this case, because the set of definitions you've read are likely not written in a way to make sense where the domain of a function is just a point. — user204305, Jan 08 '15 at 05:52

Adam Hughes · Accepted Answer · 2015-01-08T00:06:41.127

12

Technically, yes. The domain is a singleton, so the only topology on it is the trivial one, $\tau_0$. Then if $(Y,\tau)$ is any other topological space and $*$ is the one-point space

$$f:(*,\tau_0)\to (Y,\tau)$$

is trivially continuous because if $U\in\tau$ is open

$$f^{-1}(U)\in\{\varnothing, *\}=\tau_0$$

depending on whether or not $f(*)\in U$ or not.

Just to address GitGud's objection, we note that this agrees with the naïve, calculus notion of a limit as well, we're required that all $x\in \{0\}\setminus \{0\}=\varnothing$ satisfy some property. Of course there are no counterexamples (the set is empty!) so it's true, because the definition of "truth" is "that for which no instance is false." There aren't any false instances in the null set (there aren't any at all!)

This in fact follows from the above, but for the student who may not be familiar with the modern definition of continuity (which certainly resolves this problem much more directly) this is another way to think of it.

And of course, even in calculus we routinely talk about continuity on non-open subsets of $\Bbb R$. The mean-value theorem, a hallmark of calculus, has as a hypothesis "continuous on $[a,b]$." If we are not willing to extend to singletons, this seems rather unfairly favoring of intervals.

edited Jan 08 '15 at 00:06

answered Jan 07 '15 at 23:41

Adam Hughes

36,777

However if the context is calculus, the very definition of continuity demands that the function be defined in an open set and in this case it doesn't make sense to ask whether a function defined in a singleton is continuous or not. – Git Gud Jan 07 '15 at 23:53
1

@GitGud Yes, but we restrict domains all the time, including restricting to closed intervals like $[a,b]$, which only makes sense if we allow subspace topologies on the domain, since $[a,c)$ or $(d,b]$ are not open as a subsets of $\Bbb R$. It's easily translated into limits, since then punctured neighborhoods are empty so we have a vacuously true statement, so the calculus definition can still be applied, it's just a $\forall x\in\varnothing$ statement. – Adam Hughes Jan 07 '15 at 23:59
@AdamHughes Vacuously true, that's the phrase I was looking for, thanks :). – GPerez Jan 08 '15 at 00:04
@AdamHughes I hadn't notice the edit to the comment. The definition of continuous function in calculus has as a requirement that the function is defined in an open set, if I give you a function whose domain in closed (except for those cases where continuity is defined in a canonical way so that things work right, - e.g. a functions is continuous on $[a,b]$ if it is so on $]a,b[$ and the appropriate lateral limits exist), any logical analysis you do is meaningless. – Git Gud Jan 08 '15 at 00:08
@GitGud I respect that this is usually how it's done. I think it's just my personal view that the reason we don't talk about other such sets is because--under normal (i.e. calculus classroom) circumstances--we only ever talk about intervals. But I don't think that precludes the ability to talk about them in a calculus context. Even one-sided limits require $\forall x\in S$ statements for some small interval on one side of the endpoint, it seems prejudicial to preclude this for other sets when the only reason it isn't seen seems to be because it's not necessary for the exercises in the class. – Adam Hughes Jan 08 '15 at 00:11
Addressing the edit to the answer, I may even agree that not extending the definition to singletons is unfair to them. But I'm just pointing out that that's how things are usually done in calculus books, it doesn't work. I didn't write those books, it's just how it is. – Git Gud Jan 08 '15 at 00:12
@GitGud Additionally, since the op specifically requests talking about a singleton, it seems natural that he has a specific interest in how to generalize to that context, or how it can be resolved, and that the limitation comes more from how the old definition is applied in the context of his class, not how applicable the actual definition is. – Adam Hughes Jan 08 '15 at 00:13
@GitGud Re: your last comment: certainly that's fair, I agree. My last comment was written before I read your last comment, but I think it clarifies the rest of why I think it's OK in this question, here as it stands on MSE. – Adam Hughes Jan 08 '15 at 00:15
What about differentiability? – Alice Ryhl Jan 08 '15 at 06:33
@KristofferRyhl the differentiability is even more trivial. It's not even worth considering, and has no real meaning worthy of exploration. – Adam Hughes Jan 08 '15 at 06:45
The appropriate context for judging differentiability would be to consider a point as a 0-dimensional submanifold of $\mathbb{R}$. In this case a chart would map the point to a zero-dimensional vector space, where every mapping would have to be differentiable. So yes, such a function would be differentiable. – user204305 Jan 08 '15 at 07:34
@user204305 In that case, what is $f'(0)$ of my function $f:{0}\to{0},f(x)=0$ – Alice Ryhl Jan 08 '15 at 12:34
Well, it's not that simple. It would be a (the) linear mapping from the tangent space, which would be a separate copy of ${0}$, to $\mathbb{R}$. So not a number, but a mapping. With these definitions, if $f(x) = x^2$ from $(2,5)$ to $\mathbb{R}$, then $f'(3)$ is the mapping $h \mapsto 6h$ from $\mathbb{R}$ to $\mathbb{R}$. (Actually vector spaces canonically isomorphic to $\mathbb{R}$.) – user204305 Jan 08 '15 at 14:16
@KristofferRyhl it's just the same as the map itself. The derivative is an element of the zero-dimensional vector space which is linear transformations of $\Bbb R^0$, so it is just the $0$ transformation, because that is literally the only choice. – Adam Hughes Jan 08 '15 at 21:47
@AdamHughes Okay, so zero. Can that be proven from the limit definition of a derivative? – Alice Ryhl Jan 08 '15 at 21:59
@KristofferRyhl Yes, but it's stupid, as I said before: you have a vacuous limit so it's just the limit of nothing which is just $0$ much like the empty sum is $0$. It's immensely unsatisfying, and even with the more advanced definition, trivial. – Adam Hughes Jan 08 '15 at 22:16
@AdamHughes I was giving the example of a mapping defined on an open interval $(2,5)$. You seem to have misunderstood my example. I was giving a one-dimensional example to illustrate that the derivative is not just a number. – user204305 Jan 08 '15 at 23:11
@KristofferRyhl In this case, the tangent space is the vector space ${0}$, so the derivative will be the only linear mapping out of that to $\mathbb{R}$, which is the zero mapping. – user204305 Jan 08 '15 at 23:12
@user204305 ah yes, I see now, that makes more sense. I withdraw my objections. :-)
At the same time, it's clear the op already knows how to compute slopes in the calculus setting, hence my confusion.
– Adam Hughes Jan 08 '15 at 23:13

score 3 · Answer 2 · answered Jan 08 '15 at 00:01

3

Adam's answer is the most efficient way of looking at it, but here's another if you're interested.

You define continuity as $\lim\limits_{x_0\to x}f(x_0)=f(x)$ . Let's look at what this says: $$\forall\varepsilon > 0 \,\exists \delta \text{ s.t.} \forall x \in(x_0-\delta,x_0+\delta)\cap A\setminus\{x_0\}\, , \,\mid f(x_0) - f(x)\mid < \varepsilon$$

But, as you'll notice, $(x_0-\delta,x_0+\delta)\cap A\setminus\{x_0\} = \emptyset$, for any $\delta$. Now, the following is a bit of logic trickery, but it's like this: The hypothesis is false, so the implication statement is true. Every point in $(x_0-\delta,x_0+\delta)\cap A\setminus\{x_0\}$ satisfies that $\mid f(x_0) - f(x)\mid < \varepsilon $, because there are no such points to consider. It's something that took me quite some time to wrap my head around, but it's something one gets used to after proving many pathologically trivial statements.

answered Jan 08 '15 at 00:01

GPerez

6,766

No, asI said in the comment to Adam's answer, you need to start with a function defined in an open set. This is not the case. The logical analysis is senseless because of the lack of this requirement. – Git Gud Jan 08 '15 at 00:03
1

@GitGud It's more probable that you're right than wrong. However I myself never received such a definition of continuity as you state. This is all coming from an honest calculus education, in which no requirement of openness was mentioned. Spain I guess. Anyhow, it's a simple but nice exercise in logic, and OP is exposed to both sides of the discussion. I think we can be satisfied, do you not? – GPerez Jan 08 '15 at 00:14
I am satisfied because I've made my point. But I find it very strange that the domain being open wasn't a requirement for you. I've never seen it not done like that, unfortunately. – Git Gud Jan 08 '15 at 00:28
@GitGud Of course it would have to be, for certain topics. Pointwise continuity, though, wasn't one of them. – GPerez Jan 08 '15 at 02:47
Hey, so, have you concluded that the function mentioned by OP is continuous? – aarbee Jun 17 '22 at 17:55
I am writing to express my doubts of this answer's correctness. If $(_0 − \delta, _0 + \delta) \cap A ∖ {_0} = \varnothing$, then no elements can be drawn from it. As such, an element, say $x$, cannot be chosen from this set to then be used to evaluate the expression $|f(x_0) - f(x)| < \epsilon$. Without substituting the free variable $x$ in this expression, the expression remains a predicate rather than a sentence (a sentence is a formula with no free variables). ... – davidleejy Sep 11 '23 at 09:13
... As I understand, generally, predicates do not possess a truth value (cannot be true or false) and require their free variables to be substituted for more concrete ones like proper names or bounded variables to turn into a proposition which can then be assigned a truth value. For example, a predicate could be like $P(x)$ = x is blue. Without substituting x for something more concrete, say $x$:= sky, one cannot assign a truth value to $P(x)$. In this case, $|f(x_0) - f(x)| < \epsilon$ appears to remain a predicate since the $x$ cannot be furnished from an empty set, ... – davidleejy Sep 11 '23 at 09:22
... and as such, cannot have a truth value. The entire statement "$\forall \epsilon > 0, \exists \delta $ ... $|f(x_0) - f(x)| < \epsilon$" thus remains devoid of truth value. It is neither true nor false. That is, it is also not vacuously/trivially true. – davidleejy Sep 11 '23 at 09:26

Is a function defined at a single point continuous?

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