Set of increasing functions closed in the product topology of $\mathbb{R}^I$ 
Let $I$ be the unit interval and consider $\Bbb R^I$ with the product topology. Is the set $\{f \in \Bbb R^I \mid f \text{ increasing } \}$ a closed set in $\Bbb R^I$?

Is there a pictorial way to see this? The complement is the set $\{f \in \Bbb R^I \mid \exists\ x<y \text{ s.t } f(x) > f(y)\}$ so if $f \in A^c$ and $A^c$ is to be open there would need to exist $\bigcap_{x \in F} \pi^{-1}_x(U_x)$ where $F$ is a finite subset of $I$ and $\pi_x$ is the projection.
The way I've been picturing the product space with the product topology is the following

here the vertical lines are copies of $\Bbb R$ and the small vertical lines denote the sets $V_x$ and so the set $\bigcap_{x \in F}\pi_x^{-1}(V_x)$ is the set of all functions passing through these small vertical lines.
So I am looking to see if there is an open set of this form whenever there is points $x < y$ for which $f(x) > f(y)$?
 A: The convergence in $\mathbb R^I$ is the pointwise convergence. Knowing this, consider $f_n(x)=x^n$ on $I=[0,1]$ for $n\ge 1$. Then $f_n$ is increasing for every $n\ge 1$ but $f=\lim_{n\to\infty}f_n$ is the function $f(x)=0$ for $x\in [0,1)$ and $f(1)=1$. Therefore, $f$ is not increasing.
We found a sequence of elements of $A=\left\{ f\in \mathbb R^I \middle| f \text{ increasing}\right\}$ whose limit is not in $A$, so $A$ is not closed.
A: About non-decreasing question, consider the projections $p_x\colon \mathbb{R}^I\to \mathbb{R}$ sending $p_{x}(f):=f(x)$.
Of course $p_x$ is continuos for every $x\in I$, by definition of product topology. Thus is continuos also the product functions $p_x\times p_y\colon \mathbb{R}^I\to \mathbb{R}^2$ sending $f$ to $(f(x),f(y))$. Consider the following closed set $\Delta^\geq:= \{ (x,y): x\geq y\}$ of $\mathbb{R}^2$ and open set $\Delta^>:=\{\{ (x,y): x> y\}$.
Then the set of functions $S_{N-D}$ that are non-decreasing is
$$S_{N-D}=\cap_{x_0\in I}\cap_{x_1\geq x_0}\left(p_{x_1}\times p_{x_0}\right)^{-1}(\Delta^\geq)$$
that is closed.
I think that here you can just see with your eyes which is the problem for the increasing functions. In fact the set $S_I$ of the  increasing functions is
$$S_{I}=\cap_{x_0\in I}\cap_{x_1\geq x_0}\left(p_{x_1}\times p_{x_0}\right)^{-1}(\Delta^>)$$
However $\Delta^>$ is not closed, so it makes sense that the set is not necessarily closed. In fact Taladris answer shows you a counter-example.
I think we can say something more! How we would have (it’s just an intuition), we can expect that product topology is so nice that the closure of the set of increasing functions is exactly the closed set of non-decreasing functions.

$$cl(S_I)=S_{N-D}.$$
proof. Of course
$$S_{I}=\cap_{x_0\in I}\cap_{x_1\geq x_0}\left(p_{x_1}\times p_{x_0}\right)^{-1}(\Delta^>)\subseteq \cap_{x_0\in I}\cap_{x_1\geq x_0}\left(p_{x_1}\times p_{x_0}\right)^{-1}(\Delta^\geq)=S_{N-D}$$ but the right set is closed, so by closure property we have
$$cl(S_I)\subseteq S_{N-D}.$$ Let $f\in S_{N-D}$. An open neighborhood of $f$ in the product topology is the Union of a finite intersection of neighbourhoods $p_x^{-1}((-\epsilon +f(x), f(x)+\epsilon))$. Thus, for proving that $f$ is in the boundary of $S_I$ is sufficient to prove that any open neighborhood $U:=\cap_{i=1}^k p_{x_i}^{-1}((-\epsilon_i +f(x), f(x)+\epsilon_i))$ intersects always $S_I$. Let $\epsilon:= min(\epsilon_i)$ and define $F\in \mathbb{R}^I$ such that $F(x):=f(x)+\frac{\epsilon}{2}x$. By construction $F$ belongs to the neighborhood $U$ of $f$ and is also an increasing function, that means $F\in S_I\cap U$. So $f\in cl(S_I)$ and we are done.

