Continuity and sequential continuity

Question

The function $f:(X,d)\rightarrow(Y,\rho)$ is continuous if and only if $f$ is sequentially continuous (that means $x_n\rightarrow x \Rightarrow f(x_n)\rightarrow f(x)$)

Proof. First I show that if $f$ is continuous then $f$ is sequentially continuous. I consider the sequence $x_n\rightarrow x_0$ so I can find for $\varepsilon =\delta$ a value $n_\delta$ such that $|x_n-x_0|<\delta$. Doing so I can use the hypothesis that $f$ is continuous so $|f(x_n)-f(x_0)|<\varepsilon$. Now, I show the opposite implication. I know now that $$ \forall \varepsilon >0 \,\,\,\,\exists n_\varepsilon : \forall n\geq n_\varepsilon \,\,\,\,\,|x_n-x_0|<\varepsilon $$ So for a certain $\varepsilon$: $$ |f(x_n)-f(x_0)|<\varepsilon_1 $$ If I call $\varepsilon=\delta$ and $\varepsilon_1=\varepsilon$ we have the definition of continuity.

I'm not really sure, I don't know why. But can this proof be considered acceptable? I mean, it is correct and it is written in a decent way?

Answer 1

I think the forward direction is fine (even then you could probably polish it a bit better), but the backward direction is a little confusing.

I know now that $\forall \epsilon > 0, \exists n_\epsilon : \forall n \geq n_\epsilon \implies |x_n - x| < \epsilon$

So for a certain $\epsilon$:

No that isn't what you actually know. What you do know is that

$$\lim_{n \to \infty} f(x_n) = f(x)$$

and we want to prove $$\lim_{z \to x} f(z) = f(x).$$

This one says that whenever we are on a neighborhood of $x$, the images are tolerably the same. I think you were too focused on "matching symbols" to make the proof work. user117042 pretty much outlined the proof for you. I'll only comment on why he chose $\delta = 1/n$.

By choosing $\delta = 1/n$, we create a convergent sequence such that its image must also get reasonably close to its target, but that all hinges on continuity, which we assume is not true, hence the contradiction.

Also, I've noticed you aren't using the metrics in $X$ and $Y$ at all and you are simply using absolute values. You need to fix that afterwards too before you submit your answer.

Answer 2

Result: Let $f$ be a mapping from a metric space $\big( M , d \big)$ to a topological space $X$, then $f$ is continuous at $x\in M$ if and only if given any sequence $\big\{ x_n\,,\, n\in\mathbb{N}^\ast \big\}$ convergent to $x$, $f(x_n)\to f(x)$ in $X$.

Proof. Assume that $f$ is continuous at $x\in M$ and let $N$ be a neighborhood of $f(x)$, then $f^{-1}(N)$ is a neighborhood of $x$.

(Note that continuity is a topological concept.)

Since $x_n\to x$, then there exists $q\in\mathbb{N}^\ast$ such that $x_n\in f^{-1}(N)$, as $n\geq q$.

It implies that $f(x_n)\in N$ for any $n\geq q$ and thus $f(x_n)\to f(x)$.

Conversely, we assume that for any sequence $\big\{ x_n\,,\, n\in\mathbb{N}^\ast \big\}$ convergent to $x$, $f(x_n)\to f(x)$ in $X$. Suppose $f$ is not continuous at $x$, there exists some neighborhood $V$ of $f(x)$ such that $f^{-1}(V)$ is not a neighborhood of $x$, which is equivalent to $$\underline{\text{$B(x, 1/m )\cap f^{-1}(V^c) \neq \emptyset$, $\forall m\in\mathbb{N}^\ast$.}}$$

Now we can pick $x_m\in B(x, 1/m )\cap f^{-1}(V^c)$ for each $m\in\mathbb{N}^\ast$, then it is easy to observe that $d(x_m, x)\xrightarrow{m\to+\infty} 0$, implying that $f(x_m)\to f(x)$. Since $V$ is a neighborhood of $f(x)$, there exists $q\in\mathbb{N}^\ast$ such that $f(x_m)\in V$, as $m\geq q$. But $x_m\in B(x, 1/m )\cap f^{-1}(V^c)$ implies that $f(x_m)\in V^c$, $\forall m\in\mathbb{N}^\ast$, leading to a contradiction. Hence $f$ is continuous at $x$ and the proof is finished.

Q.E.D.

Attention: If $M$ is a general topological space, sequential continuity does not necessarily imply the continuity.

Answer 3

I think your proof is bad written. For the direction (=>) we can do straightforwardly. For the reverse direction, I suggest to use contradiction argument: negating the definition of continuity yields that there exists $\epsilon>0$ such that for any $\delta(\epsilon)>0$ ..Then we only need to choose $\delta=1/n$ and it then appears a sequence to deal with! (sorry for my mistake, I've corrected it)

Answer 4

FORMAL PROOF

$(\Rightarrow):$ Let $f$ be continuous at $a$ and let $(x_n)\in A^{\mathbb{N}}$ with $x_n\to a,$ and $\epsilon>0.$

$\left.\begin{array}{r} \epsilon>0 \\ \\ f \text{ is continuous at } a \end{array} \right\}\Rightarrow \begin{array}{c} \\ \\ \left. \begin{array}{r} (\exists \delta>0)(A\cap (a-\delta,a+\delta)\subseteq f^{-1}[(f(a)-\epsilon,f(a)+\epsilon)]) \\ \\ (x_n\to a)\left((x_n)\in A^{\mathbb{N}}\right) \end{array} \right\} \Rightarrow \end{array}$

$\begin{array}{r} \Rightarrow (\exists K\in\mathbb{N})(n\geq K\Rightarrow x_n\in A\cap (a-\delta,a+\delta)\subseteq f^{-1}[(f(a)-\epsilon,f(a)+\epsilon)]) \end{array}$

$\begin{array}{r} \Rightarrow (\exists K\in\mathbb{N})(n\geq K\Rightarrow f(x_n)\in f[A\cap (a-\delta,a+\delta)]\subseteq (f(a)-\epsilon,f(a)+\epsilon)) \end{array}$

$\Rightarrow (\exists K\in\mathbb{N})(n\geq K\Rightarrow f(x_n)\in (f(a)-\epsilon,f(a)+\epsilon))$

$\Rightarrow f(x_n)\to f(a).$ $-----------------------------------$

$(\Leftarrow):$ Let $f$ be not continuous at $a$.

$$f \text{ is not continuous at } a$$$$\Rightarrow$$ $$(\exists \epsilon>0)(\forall\delta >0)(f[A\cap (a-\delta,a+\delta)]\nsubseteq (f(a)-\epsilon,f(a)+\epsilon))$$ $$\Rightarrow$$ $$(\exists \epsilon>0)(\forall n\in\mathbb{N})\left(f\left[A\cap \left(a-\frac1n,a+\frac1n\right)\right]\nsubseteq (f(a)-\epsilon,f(a)+\epsilon)\right)$$ $$\Rightarrow$$ $$(\exists \epsilon>0)(\forall n\in\mathbb{N})\left(\exists x_n\in A\cap \left(a-\frac1n,a+\frac1n \right)\right)(f(x_n)\notin (f(a)-\epsilon,f(a)+\epsilon)$$ $$\Rightarrow$$ $$\left(\exists (x_n)\in A^{\mathbb{N}}\right)(x_n\to a)(f(x_n)\nrightarrow f(a)).$$

$-----------------------------------$

NOTE:

$$\left[\left(\forall (x_n)\in A^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to f(a))\right] \Rightarrow \left[f \text{ is continuous at } a\right]$$

$$\equiv$$

$$\left[f \text{ is continuous at } a\right]'\Rightarrow \left[\left(\forall (x_n)\in A^{\mathbb{N}}\right)(x_n\to a\Rightarrow f(x_n)\to f(a))\right]'$$

$$\equiv$$

$$f \text{ is not continuous at } a\Rightarrow \left(\exists (x_n)\in A^{\mathbb{N}}\right)(x_n\to a\wedge f(x_n)\nrightarrow f(a)).$$