Let $B(x)$ be a Brownian motion. We know by definition that $B(x+1) - B(x)$ ~ $N(0,1)$. I am trying to understand if that means that $B(x+1)-B(x) = PDF(x)$, where PDF(x) is the probability density function of $N(0,1)$ at the point $x$.
Can you tell me if its right? and if it isnt, what is the actual meaning of $B(x+1)-B(x)$ ~ $N(0,1)$?
I'm not sure where the confusion is raising, but I'll try.
If we define
$$ \Delta_{x} := B(x+1) - B(x) $$
we see that this variable is normally distributed. What is this? It is the increment. This means that if at step $x$ we are at $B(x)$ we can find where we will be at $x+1$ using
$$ B(x+1) = \Delta_x + B(x) $$
but $\Delta_x$ is a random variable! This means that we can have certain probabilities of being somewhere in the next step.
For simplicity, let's image that at $x=0$ we start from $B(0)=0$. It means that
$$ B(1) = \Delta_0 $$
which is a normally distributed random variable! The gaussian probability tells you that the probability density function of $B(1)$ is gaussian. So, in this case, you will have a high probability of being around $0$ at the next step and a lower probability of being far from it.
I've used your notation here, but I suggest using $t$ instead of $x$ as it is more intuitive to understand that it is the time.
$B(x+1)-B(x)$ is a random variable which follows a standard normal distribution, that is, for each Borel set $A$ of the real line, $$ \mathbb P(B(x+1)-B(x)\in A)=\frac 1{\sqrt{2\pi}}\int_A\exp\left(-t^2/2\right)dt. $$
