$X \subset \mathbb{R}^m$. Let $\phi:x \to \mathbb{R}^n$ be bounded. $\phi$ is continuous $\iff$ its graph is closed. $X \subset \mathbb{R}^m$. Let $\phi:x \to \mathbb{R}^n$ be bounded. Then $\phi$ is continuous $\iff$ its graph is closed.


I was asked to prove this, but I believe it's false.

Let $Gra(\phi) = \{(x, \phi(x))|x \in X\}$ the Graph of $\phi$.

Suppose X is not closed. Then exists $a \in \bar{X}-{X}$ (being $\bar{X}$ the closure of X).

Therefore exists $x_n \in X$ s.t. $x_n \to a$. The sequence $\phi(x_n)$ has a convergent subsequence $\phi(x_{k_n})$ since it's bounded. Let $\lim \phi(x_{k_n})=b$

Then $(a,b) \in \overline{Gra(\phi)}$  (the closure of $Gra(\phi)$) but $(a,b)\not \in Gra(\phi) \implies Gra(\phi)$ is not closed.

Is this correct?
Of course, when $X$ itself is not closed in $\mathbb{R}^m$, the graph cannot be closed. We henceforth assume that $X$ is closed.

Let $\mathcal{C}(X)$ denote the set of convergent sequences $\mathbf{x}=(x_n)_{n\in\mathbb{N}}\subset X$. For any such sequence we will denote its limit by $x$. Notice that $x\in X$ by the assumption that $X$ is closed.

On the one hand, we have:

\begin{align}
\phi \text{ is continuous}&amp;\iff \phi \text{ is sequentially continuous}\\
&amp;\iff \forall \mathbf{x}\in\mathcal{C}(X), \,\phi(x)=\lim_{n\to\infty}\phi(x_n)\tag{1}\end{align}

On the other, we have

\begin{align}
Gra(\phi) \text{ is closed}&amp;\iff Gra(\phi) \text{ is sequentially closed}\\
&amp;\iff \forall \text{ convergent } \big(x_n,\phi(x_n)\big),\, \lim_{n\to\infty}\big(x_n,\phi(x_n)\big) \in Gra(\phi)\tag{2}
\end{align}

It is clear that $(1)\implies (2)$.
Indeed, if $\big(x_n,\phi(x_n)\big)$ is convergent, then $\mathbf{x}=(x_n)$ is convergent and hence by $(1)$ so too is $\big(\phi(x_n)\big)$,
with $\lim_{n\to\infty}\phi(x_n)=\phi(x)$.
Hence,

$$\lim_{n\to\infty}\big(x_n,\phi(x_n)\big)=\big(x,\phi(x)\big),$$

which belongs to $Gra(\phi)$ by definition.



We will show that when $\phi$ is bounded, $(2)\implies(1)$, which conludes the proof.

Suppose $\phi$ is bounded and let $\mathbf{x}\in\mathcal{C}(X)$.
Since $\phi$ is bounded and $(x_n)$ is convergent (and hence bounded), we have that $\big(\phi(x_n)\big)$ is a bounded sequence.
By the Bolzano-Weierstrass Theorem, there is a convergent subsequence $\big(\phi(x_{n_k})\big)$ of $\big(\phi(x_n)\big)$.

Of course, $(x_{n_k})$ it itself convergent and converges to $x$, the same limit of $(x_n)$.
It follows that $\big(x_{n_k},\phi(x_{n_k})\big)$ is convergent, and by $(2)$ we have that $\lim_{k\to\infty}\big(x_{n_k},\phi(x_{n_k})\big)=\big(x,\phi(x)\big)$.
In particular, $\lim_{k\to\infty}\phi(x_{n_k})=\phi(x)$.

It then suffices to show that $\big(\phi(x_n)\big)$ is convergent;
in this case, its limit must coincide with that of $\big(x_{n_k},\phi(x_{n_k})\big)$, that is, must equal $\phi(x)$. We show this by contradiction.

Indeed, if that were not the case, then there would be some $\epsilon&gt;0$ and a subsequence $(x_{m_k})$ of $(x_n)$ with $d\big(\phi(x_{m_k}),\phi(x)\big)\geq \epsilon$ for all $k$.
Now, $\big(\phi(x_{m_k})\big)$ is of course bounded, so by the Bolzano-Weierstrass Theorem it must have a convergent subsequence, say $\left(\phi\left(x_{m_{k_j}}\right)\right)$.
By construction, we have that $$y=\lim_{j\to\infty}\phi\left(x_{m_{k_j}}\right)\neq \phi(x)$$

But $\left(x_{m_{k_j}}\right)$ is a subsequence of $(x_n)$, and hence converges to $x$.
It follows that $$\lim_{j\to\infty}\left(x_{m_{k_j}},\phi\left(x_{m_{k_j}}\right)\right)=(x,y)$$
and hence does not belong to $Gra(\phi)$, in contradiction with $(2)$. $\square$.




  Notice that in the proof above we didn't actually need the hypothesis that $\phi$ be bounded, but rather that $\phi$ 'preserves boundedness', ie, $\phi$ takes bounded sets to bounded sets.
This is wrong (for unbounded function). take $\phi : [0, \infty] \rightarrow R$ with $\phi(x) = \frac{1}{x}$ for all $x \in (0, \infty)$ and $\phi (0)=0$. its graph is closed but $\phi$ is not continuous!

EDIT: The above example is for unbounded function on limited domain (before editing the original question). If we assume function is bounded then the claim  is correct.

Proof: For Left to Right: the function $F(x) = (x, \phi(x))$  is continuous (now to show graph is closed in $X \times R^n$, use the sequential definition of continuity of $F$).
For Right to Left: similar to your argument, take $x_n \rightarrow x \in X$. want to show that $\phi(x_n) \rightarrow \phi(x),$ if actually $\phi(x_n) \nrightarrow \phi(x)$ then (since $\phi$  is bounded)  $\phi (x_n)$ has a convergent subsequence, say $\phi (x_{n_{k}}) \rightarrow y \neq \phi(x)$, i.e., $$(x_{n_{k}},  \phi (x_{n_{k}}) ) \rightarrow (x,y)\notin \text{Graph}.$$

Which is contradicting with Graph being Closed.