neighbourhood concept-how come $[0,1]^2$ is not a neighbourhood of $(1,1)$ Below is the capture from lecture slides
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I know what the neighbourhood definition is, and it seems to me $[0,1]^2$ is the neighbourhood of $(1,1)$ and $(0,\frac{1}{2})$
Further edited
Thank you so much for your reply. I regard $[0,1]^2$ as a unit square if I do not get it wrong, those examples are at the boundary of the set. I just noticed closed set is not a neighbourhood of its end-point This bugs me!!
According to this, I believe the set is not the neighbourhood of $(0,0)$ as well, but Isn't  $[0,\frac{1}{2})$ open in $[0,1]$? This is analogy that I rely on to make the claim.

When the slide says that a neighborhood should surround the point, it means that there is a disk of positive radius centered at the point that is included in the neighborhood. $[0,1]\times [0,1]$ is not a neighborhood of $(1,1)$ because you can escape starting from $(1,1)$ following the diagonal $(t,t)$, $t > 1$ without meeting $[0,1]\times [0,1]$.
Note that we are talking of neighborhoods of points in the plane. The notion of open sets and neighborhood are relative to the space the point belongs to.
A subset is open or not in a given set. For example, $[0,1/2)$ is an open subset of $[0,1]$ but not an open subset of ${\bf R}$.
In the slides, the notion of neighborhood and open sets are relative to the plane itself. If instead of the plane, you consider the space $[0,1]\times [0,1]$, then the subset $(1/2,1] \times (1/2,1]$ is a neighborhood of the point $(1,1)$ in the square $[0,1] \times [0,1]$.
