6

Let $f(x)=\lfloor x \rfloor $ and imagine posing the following questions.

  1. Is $f(x)$ continuous at $x=0$?
  2. Is $f(x)$ continuous on $[0,1)$?

For the first question, since $\displaystyle \lim_{x\rightarrow 0} f(x)$ does not exist, we must answer no.

For the second question, since $\displaystyle \lim_{x\rightarrow 0^+} f(x)$ exists and $\forall a \in (0,1) : \displaystyle \lim_{x\rightarrow a} f(x) = f(a)$, we should answer yes.

These are the answers to these two questions based on my understanding of what it means to be continuous at a point and what it means to be continuous on a(n) (closed) interval.

However, in retrospect, this seems bizarre to me given that we are saying that $f(x)$ is not continuous at $0$, while $f(x)$ is continuous on $[0,1)$ even though $0 \in [0,1)$. Is this really the case?

Trevor Kafka
  • 1,280

2 Answers2

11

The reason for the seeming bizarritude is simply that you are suppressing the domains of the functions at issue, and combining that with ambiguous wording. Once you pay attention to domains and use unambiguous language, the problem goes away.

In the first instance, you are considering the function $f$ with domain $\mathbb R$ and asking if that function $f : \mathbb R \to \mathbb R$ is continuous at $0$, and the answer is no.

In the second instance, you are restricting $f$ to have domain $[0,1)$, and asking if that restricted function $f : [0,1) \to \mathbb R$ is continuous at $0$, and the answer is yes.

Lee Mosher
  • 135,265
  • Is it not appropriate to say $f:\mathbb R \rightarrow \mathbb R : x \mapsto\lfloor x \rfloor$ is continuous on $[0,1)$? – Trevor Kafka Sep 22 '22 at 19:39
  • 2
    Appropriateness aside, you tell me what that ambiguous sentence means and I'll tell you whether it is true or false. If it means "The function $f : \mathbb R \to \mathbb R$ is continuous at every point of $[0,1)$" then it is false. If it means "The function obtained from $f : \mathbb R \to \mathbb R$ by restricting the domain to $[0,1)$ is continuous" then it is true. – Lee Mosher Sep 22 '22 at 23:42
  • Lee, a less passive-agressive response would have been something like:

    "Not quite, but you could say ______ or _______ to be precise."

    Thank for your input, but no thanks for the attitude.

     – Trevor Kafka Sep 24 '22 at 23:12
  • 2
    Sorry it didn't work out for you, but my intentions are not passive aggressive. My intentions are to teach mathematical thinking, and in this instance to teach how to root out one's own logical imprecisions and ambiguities. – Lee Mosher Sep 25 '22 at 12:57
2

This might be easier to understand using the open set definition of continuity; i.e., a function $f(x) : X \to Y$, where $X$ and $Y$ are topological spaces endowed with specified topologies, is continuous if $f^{-1}(U)$ is open in $X$ for every open set $U$ in $Y$. The upshot here is that continuity of a function is dependent on its domain. In the first case, the domain of $f$ is $\mathbb{R}$, while in the second case it is $[0, 1)$. In $[0, 1)$ with the subspace topology, sets of the form $[0, a)$, $a < 1$ are open.

user1090793
  • 1,491