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Let $f \colon [a,b] \to \mathbb{R}$, and for $x \in [a,b]$, define $$L(x) = \{y \in \mathbb{R}: \text{there exists a sequence} \left(x_n\right) \text{with } x_n \to x \text{ and } f(x_n) \to y\}.$$ In other words, $L(x)$ is the set of all the different 'possible' values for $\lim_{t \to x}f(t)$ depending on how we approach $x$ (of course, if the limit exists, then that will be the only element of $L(x)$).

We wish to show that $L(x)$ is closed, so take $y_r \in L(x)$ for $r \geq 1$, and suppose the sequence $y_r \to y$. To show that $y \in L(x)$, trying to directly construct a suitable sequence to meet the definition seems difficult, as there's no obvious way to 'glue together' the different sequences for each of the $y_r$s. One idea I had was the diagonal sequence, i.e. let $\left(x_{r,k}\right)_{k \geq 1}$ be the sequence we get from the definition of $L(x)$ for each $y_r$, and consider the new sequence $\left(x_{n,n}\right)_{n \geq 1}$ - but there doesn't seem to be an obvious way to show this even converges to $x$. A proof by contradiction also doesn't seem to work. Thanks in advance for any thoughts!

Prasiortle
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    Some relevant questions for this can be found in this google search. However, most of these that ask the question you're asking are for a metric space setting, and many deal with the notion for a sequence (which is really just a specific type of function). Some additional questions about this notion can be found in this other google search. – Dave L. Renfro Dec 31 '23 at 08:50
  • Obviously if there is a general argument for all metric spaces, then all the better! The closest question I could find on here was https://math.stackexchange.com/questions/3898041/limit-set-of-a-function-is-a-closed-set, but the definition of the limit set in that one was a bit different (it considered all points in the domain of the function, rather than one specific point $x$), and the argument seemed rather indigestible. – Prasiortle Dec 31 '23 at 09:22
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    Obviously if there is a general argument for all metric spaces, then all the better! --- Since you didn't discuss you math background, I thought I should mention that many were in a metric space context, since if you don't know anything about metric spaces (e.g. maybe at the beginning of studying very basic real analysis, or maybe this is from an honors elementary calculus course) then those questions wouldn't be of much help to you. – Dave L. Renfro Dec 31 '23 at 09:54
  • Yeah that's fine lol; as you'll see from some of my previous questions, I'm familiar with real and complex analysis and topology, and to some extent with basic measure theory. The source I happened to get this problem from was a first-year real analysis exam, hence the more basic formulation. – Prasiortle Dec 31 '23 at 09:57
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    In case anyone thinks this is a duplicate, make sure that the cited duplicate is for real valued functions of a real variable, and it doesn't involve metric or topological spaces or ${\mathbb R}^n$ for $n \geq 2$ (or even $\mathbb C$ or ${\mathbb R}^2).$ I think it would be useful to have a nicely written answer for just this context, since the result is definitely understandable by someone who hasn't been exposed to these more general settings (e.g. someone reading Spivak's Calculus who has not yet dealt with functions of more than one variable). – Dave L. Renfro Dec 31 '23 at 10:26
  • If $y_r \to y$ and there is a sequence of sequences $(x_n)^r$ for each $r$ such that $f(x_r) \to f(x)$, then isn't it somewhat trivial (using the usual triangle inequality) to show that we can pick elements $x_i$ from the sequence of sequences $(x_n)^r$ such that $f(x_i)$ gets arbitrarily close to $y$? – TryingHardToBecomeAGoodPrSlvr Dec 31 '23 at 12:47
  • Starting with $\epsilon > 0$, we get $\left\lvert y_r-y\right\rvert < \epsilon$ for $r \geq R$ and $\left\lvert f\left(x_{r,n}\right)-y_r\right\rvert < \epsilon$ for $n \geq N$, so the triangle inequality indeed gives $\left\lvert f\left(x_{r,n}\right)-y\right\rvert < 2\epsilon$ for $n \geq N, r \geq R$. But that's exactly where I got stuck before - this sequence converging to $x$ depends on $\epsilon$, whereas we need to define a single sequence converging to $x$ and then show it satisfies the $\epsilon\text{-}N$ definition of convergence ... 1/ – Prasiortle Dec 31 '23 at 13:07
  • ... The answers below show how to deal with this, by ignoring $\left\lvert y_r-y\right\rvert$, replacing $\epsilon$ with $1/r$, and then using the squeeze theorem. 2/2 – Prasiortle Dec 31 '23 at 13:07

2 Answers2

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I have not seen the edit of the other answer while typing on my phone. The argument is essentially the same. I'll keep my answer in the hope that the slightly different phrasing might help somebody.

Let $(y_n)_n\subset L(x)$ be a sequence converging to $y\in \mathbb{R}$. We need to show that $y\in L(x)$.

For every $y_n$ we have a sequence $(x_m^{(n)})_m \subset [a,b]$ such that $\lim_{m\rightarrow \infty} x^{(n)}_m=x$ and $\lim_{m\rightarrow \infty } f(x^{(n)}_m)=y_n$. Thus, we can pick $k(n)\in \mathbb{N}$ such that $\vert x^{(n)}_{k(n)}-x\vert <1/n$ and $\vert f(x^{(n)}_{k(n)})- y_n\vert<1/n$.

First we note that $\lim_{n\rightarrow \infty } x^{(n)}_{k(n)} = x$. Furthermore, $$ \vert y-f(x^{(n)}_{k(n)})\vert \leq \vert y-y_n\vert + 1/n$$ and therefore $$\lim_{n\rightarrow \infty} f(x^{(n)}_{k(n)}) = y.$$ Thus, $y\in L(x)$ and $L(x)$ is closed.

The same argument works for general metric spaces, one just needs to replace the absolute value by the corresponding expression involving the metric.

Severin Schraven
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  • Thanks a lot! I had indeed tried to argue $\left\lvert f(x_n)-y\right\rvert \leq \left\lvert f(x_n)-y_n\right\rvert + \left\lvert y_n-y\right\rvert$, but I didn't see that you could use $n$ itself as a sort of global synchonising variable for the different sequences $(x_m)$ and then eliminate the resulting $1/n$ using the squeeze theorem. – Prasiortle Dec 31 '23 at 12:47
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    @Prasiortle You're welcome. As I wrote above, the technique is practically identical as the one Asigan used in his answer. They just prove a more powerful characterization first and then deduce the claim from this stronger result. It's a bit unfortunate that I had not seen their edit before. – Severin Schraven Dec 31 '23 at 13:01
  • Indeed - but I still like your argument as it directly constructs the sequence converging to $x$ rather than going by contradiction. – Prasiortle Dec 31 '23 at 13:03
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Remark: In the following $L(x)$ is understood as \begin{equation} L(x) = \{y \in \mathbb{R}: \text{exists a sequence} \{x_n\}\text{s.t. } x_n \to x, x_n\ne x\ \forall n, \text{ and } f(x_n) \to y\}. \end{equation}

We call this "narrow $L(x)$". If $L(x)$ is understood as \begin{align} L(x) = &\{y \in \mathbb{R}: \text{exists a sequence} \{x_n\}\text{s.t. } x_n \to x,\\ & \underline{\text{we do not require }} "x_n\ne x\ \forall n", \text{ and } f(x_n) \to y\}, \end{align} then this more general $L(x)$ is the union of $\{f(x)\}$ and the narrow $L(x)$, since if $x_n\equiv x$ for all $n$ then $f(x_n)\equiv f(x)$. It is immediate that closedness of narrow $L(x)$ implies closedness of general $L(x)$, since the latter is the union of narrow $L(x)$ (closed) and one-point set $\{f(x)\}$.

We ${\color{violet} {\text{claim}}}$ $u\notin L(x)$ $\iff$ exists $(u_1,u_2)\ni u$ a neighbourhood of $u$, and $(c,d)\cap [a,b]$ a neighbourhood of $x$ in $[a,b]$, s.t. the set \begin{equation*} \{y\in (c,d)\cap [a,b]-\{x\}: f(y)\in (u_1,u_2) \} \end{equation*} is finite. (The proof is added at the end of the answer).

Therefore any $v\in (u_1,u_2)$ is neither an element of $L(x)$ (Apply the above statement to ${v}\in (u_1,u_2)$ and $(c,d)\cap [a,b]$). Therefore $\mathbb{R}-L(x)$ is open, i.e. $L(x)$ is closed.

Proof of the ${\color{violet} {\text{claim}}}$:

($\implies$): Assume for all $(u_1,u_2)\ni u$ and $(c,d)\cap [a,b]$, the set \begin{equation*} \{y\in (c,d)\cap [a,b]-\{x\}: f(y)\in (u_1,u_2) \} \end{equation*} is infinite. Then it is a fortiori non-empty. Take $(u_1,u_2)=\left(u-\frac{1}{n},u+\frac{1}{n}\right)$ and $(c,d)=\left(x-\frac{1}{n},x+\frac{1}{n}\right)$ and select a real number in \begin{equation*} \left\{y\in \left(x-\frac{1}{n},x+\frac{1}{n}\right)\cap [a,b]-\{x\}: f(y)\in \left(u-\frac{1}{n},u+\frac{1}{n}\right) \right\}. \end{equation*} Denote the real number we select corresponding to $n$ by $x_n$. Then \begin{equation*} \begin{cases} \mid x_n-x \mid<\frac{1}{n}\\ \mid f(x_n)-u \mid <\frac{1}{n}. \end{cases} \end{equation*} Therefore $x_n\to x$ and $f(x_n)\to u$, which contradicts the hypothesis.

($\impliedby$): Since there are finitely many $y\in (c,d)\cap [a,b]-\{x\}$ s.t. $f(y)\in (u_1,u_2)$, we can construct a smaller interval $(c_1,d_1)$ contained in $(c,d)$ s.t. \begin{equation*} \{y\in (c_1,d_1)\cap [a,b]-\{x\}: f(y)\in (u_1,u_2) \}=\varnothing. \end{equation*} Then for all $\{x_n\}$ converging to $x$, exists $N$ s.t. $x_n\in (c_1,d_1)\cap [a,b]-\{x\}$ for all $n>N$. But for all $n>N$, since $x_n\in (c_1,d_1)\cap [a,b]-\{x\}$, we must have $f(x_n)\notin (u_1,u_2)$. Therefore $f(x_n)\not\to u$. Since $\{x_n\}$ is an arbitrary sequence converging to $x$, we must have $u\notin L(x)$.

Asigan
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    Not sure how you get that first implication - could you add some more details? – Prasiortle Dec 31 '23 at 11:17
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    @Prasiortle I have added the proof. And there was a confusing typo: "$L(x)$" should be "$(u_1,u_2)$". So I should claim the responsibility for the original answer being confusing. You know I have first posted a ridiculously wrong answer. I was hurrying to write a correct one (and it would soon be dinner time in my region while writing the answe :) ), so I confess I did not write this answer patiently at first. I apologize for the confusion I caused. – Asigan Dec 31 '23 at 12:26
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    Thanks so much - I was going to say, your original definition of the set didn't seem to involve $u$ or $(u_1,u_2)$ at all! One might also wonder whether there's a proof using the sequential definition of a closed set rather than the open complement definition, but it seems like the other answer has provided just that, so I'm happy to make yours the accepted answer. – Prasiortle Dec 31 '23 at 12:39
  • @Prasiortle I suddenly realized whether one allow $x_n$ take the value of $x$ may cause a little trouble. You ca see my new edit, which added a remark at the opening of the answer. – Asigan Dec 31 '23 at 13:29
  • I can't see why your original argument wouldn't have worked for a sequence with $x_n = x$... – Prasiortle Dec 31 '23 at 13:43
  • @Prasiortle If my claim is understood by the way that in the definition of $L(x)$, $x_n$ can be $x$, then consider $f:[-1,1]\to\mathbb{R}$. $f(0)=1$ and $f(x)=0$ whenever $x\in [-1,1]-{0}$. We have $1\in L(0)$ since $1=f(0)$. But exists a neighbourhood of $0$, say, $(-0.1,0.1)$ and a neighbourhood of $f(0)=1$, say, $(0.999,1.001)$ s.t. \begin{equation} {y\in (-0.1,0.1): f(y)\in (0.999,1.001)}={0} \end{equation} is finite, but $1\in L(0)$. Therefore for the claim to hold, the $L(x)$ must be understood in the narrow sense. – Asigan Dec 31 '23 at 14:13
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    Got it, thanks! We can still define an interval $(c_1,d_1)$, but then it's no longer true that "exists $N$ s.t. $x_n \in (c_1,d_1) \cap [a,b]−{x}$ for all $n>N$". – Prasiortle Dec 31 '23 at 14:24
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    @Prasiortle Very true. At first in the proof $(\impliedby)$ I wrote exists $c_1, \ d_1$ s.t. \begin{equation} {y\in (c_1,d_1)\cap [a,b]: f(y)\in (u_1,u_2) }=\varnothing. \end{equation} This can not hold if $u=f(x)$. – Asigan Dec 31 '23 at 15:10