Proof that $f : \mathbb{R} \longrightarrow \mathbb{R}$ is continuous iff $f(\overline{X}) \subset \overline{f(X)}$ using $\varepsilon-\delta$.

Question

I'm helping an acquaintance to prove this statement, but I'm having doubts if the proof I did is correct. Can someone help me and tell me if this is correct?

I'm in doubt on ($\Leftarrow$).

$\Rightarrow$ Suppose f continuous. Let $X \subset \mathbb{R}$ and $a \in \overline{X}$. Because $f$ is continuous, given $\varepsilon > 0$, exists $\delta > 0$ such that

$$\mid z - a \mid \ < \ \delta \ \Rightarrow \ \mid f(z) - f(a) \mid \ < \ \varepsilon.$$

How $a \in \overline{X}$, we have that

$$[(a - \delta, a + \delta) - \{a\} ]\cap X \ne \emptyset. $$

Then there exists $x \in X$ such that

$$\mid x - a \mid \ < \ \delta \ \Rightarrow \ \mid f(x) - f(a) \mid \ < \ \varepsilon$$

$$\Leftrightarrow f(x) \in (f(a) - \varepsilon, f(a) + \varepsilon). $$

How $\varepsilon$ is arbitrary, we have that $f(a) \in \overline{f(X)}$.

$\Leftarrow$ Suppose that $f$ is discontinuous in a point $a \in \mathbb{R}$. Then there exists $\varepsilon > 0$ such that for every $\delta > 0$, we have that

$$\mid z - a \mid \ < \ \delta \ \Rightarrow \ \mid f(z) - f(a) \mid \ \geq \ \varepsilon.$$

Given $X \subset \mathbb{R}$ and $a \in \overline{X}$, we have that $f(a) \in \overline{f(X)}$. Therefore

$$[(f(a) - \varepsilon, f(a) + \varepsilon) - \{a\}] \cap f(X) \ne \emptyset.$$

Then exists $x \in X - \{a\}$ such that

$$\mid x - a \mid \ < \ \delta \ \Rightarrow \ \mid f(x) - f(a) \mid \ < \ \varepsilon,$$

what is a contradiction.

I don't know if my loop is correct as I take a set X where a is an accumulation point. In the book in which I study and the solutions to the questions I see, this is not used.

@FShrike I know this proof in the context of general topology. It so happens that at the moment I am dealing with real analysis and I would not like to use this more general definition of continuity, because despite being equivalent, I do not know if the professor would accept it. — Gleberson Antunes, Jul 07 '23 at 12:22
Does this answer your question? A map is continuous if and only if for every set, the image of closure is contained in the closure of image — Angelo, Jul 07 '23 at 12:35
@GlebersonAntunes, this is not a “do my homework” website. Moreover your question is a duplicate, so please delete it. — Angelo, Jul 07 '23 at 12:38
To be fair they seem to want an $\epsilon$-$\delta$ proof so this might not quite be a duplicate — FShrike, Jul 07 '23 at 13:09
I personally like this question better because it discusses lots of different ways that images and preimages interact with closures and interiors, and discusses them all. @FShrike — Jakobian, Jul 07 '23 at 14:10

score 0 · Answer 1 · answered Jul 07 '23 at 14:28

Suppose that $f$ is discontinuous in a point $a \in \mathbb{R}$. Then there exists $\varepsilon > 0$ such that for every $\delta > 0$, we have that

$$\mid z - a \mid \ < \ \delta \ \Rightarrow \ \mid f(z) - f(a) \mid \ \geq \ \varepsilon.$$

That's wrong. (I overlooked that at first, which is why I deleted my earlier answer.)

You only have $$ |z-a|<\delta\not\Rightarrow|f(z)-f(a)|<\epsilon, $$ which is not the same.

FShrike · Answer 2 · 2023-07-07T14:47:28.143

The question in the title is very much a duplicate, but the OP seemed to want some kind of classic real analytic proof.

Then there exists $\varepsilon > 0$ such that for every $\delta > 0$, we have that: $$|z - a|<\delta\implies|f(z) - f(a)|\ge\varepsilon$$

This is not right. Discontinuity of $f$ at $a$ means:

There exists $\epsilon>0$ such that there is no $\delta>0$ with $|z-a|<\delta\implies|f(z)-f(a)|<\epsilon$

Or equivalently:

There exists $\epsilon>0$ such that for all $\delta>0$ there exists some $x$ with $|x-a|<\delta$ yet $|f(x)-f(a)|\ge\epsilon$

If you want a proof with a 'real analysis feel' to it (rather than a topological one, which is admittedly far easier) then we can try this:

Let $x\in\Bbb R$ be arbitrary and $(x_n)_{n\in\Bbb N}$ a real sequence with limit $x$. Let $(n_k)_{k\in\Bbb N}$ be any increasing sequence of naturals.

Observe that for any $m\in\Bbb N$: $$\begin{align}f(x)&\in f(\{x\}\cup\{x_{n_k}:k\ge m\})\\&=f(\overline{\{x_{n_k}:k\ge m\}})\\&\subseteq\overline{f(\{x_{n_k}:k\ge m\})}\end{align}$$

It follows that there is an increasing sequence $(k_j)_{j\in\Bbb N}$ of naturals with: $$\lim_{j\to\infty}f(x_{n_{k_j}})=f(x)$$

Since $(n_k)_{k\in\Bbb N}$ was arbitrarily taken, it follows by the 'Urysohn subsequence principle' that: $$\lim_{n\to\infty}f(x_n)=f(x)$$Since $(x_n)_{n\in\Bbb N}$ was arbitrarily taken, it follows $f$ is continuous at $x$. It follows further that $f$ is continuous, since $x$ was arbitrarily taken.

You can also do an $\epsilon$-$\delta$ proof but the only way I see to do this would just be a lazy mimicking of the topological proof.

Let $a\in\Bbb R$ and $\epsilon>0$ be arbitrary and $K:=\{x\in\Bbb R:|f(x)-f(a)|\ge\epsilon\}$.

Because: $$f(\overline{K})\subseteq\overline{f(K)}\subseteq\overline{\Bbb R\setminus(a-\epsilon,a+\epsilon)}=\Bbb R\setminus(a-\epsilon,a+\epsilon)$$

It follows that if $x\in\overline{K}$ then $|f(x)-f(a)|\ge\epsilon$ too, so $\overline{K}=K$. Because $K$ is closed and $a\notin K$ I know there must be some $\delta>0$ such that $(a-\delta,a+\delta)\cap K=\emptyset$. Otherwise, $a\in\overline{K}=K$ would follow which is absurd.

But then it must be true that: $$|x-a|<\delta\implies|f(x)-f(a)|<\epsilon$$

Proof that $f : \mathbb{R} \longrightarrow \mathbb{R}$ is continuous iff $f(\overline{X}) \subset \overline{f(X)}$ using $\varepsilon-\delta$.

2 Answers2