Is boundary of a subspace strictly less dimensional then space in topology Answering this you just save my time and health. It is very bad question by  itself but very easy to answer I think. 
In topology the boundary of a set $S$ is the set of points in the closure of $S$ not belonging to the interior of $S$. 
Is it true that if dimension of a space is defined then boundary is strictly less dimensional?
I was thinking so, but stuck on the case of plane and line. 
I used to think that a line on a plane has empty boundary (or segment has endpoints as boundary) but now stuck ...
Consider a plane $\mathbb R^2$ with euclidean topology and a line $l$. 
Let $x\in l$. Arbitrary ball $B(x,r)$ with $r>0$ intersects complement of a line and hence interior of $l$ is empty (am I wrong here?).
Since $l$ is closed (every point of a complement has a ball separeting it from the line) then $cl(l)\setminus int(l)=l$. 
What do I wrong? :O    
 A: This idea has led to one of the simplests concepts to define a topological dimension, the small inductive dimension $\operatorname{ind}(X)$ of a topological space. We say that a space has $\operatorname{ind}(X) \le n$ iff $X$ has a base $\mathcal{B}$ (of open sets) such that $\operatorname{ind}(U) \le n-1$ for all $U \in \mathcal{B}$.
To start the induction, one defines $\operatorname{ind}(\emptyset)=-1$, so a space has small inductive dimension $\le 0$ if it has a base with clopen (closed-and-open) sets (these are exactly the sets with empty boundary), and includes spaces like $\Bbb Q$ and the irrationals. 
To finish off the definition $\operatorname{ind}(X) =n$ is precisely when $\operatorname{ind}(X) \le n$ holds and $\operatorname{ind}(X) \le n-1$ fails. 
There are other inductive ways to define dimension function, like the large inductive dimension $\operatorname{Ind}(X)$, and the (Lebesgue) covering dimension   $\dim(X)$, and the Brouwer "Dimensionsgrad" (fallen out of use, but of historic importance). 
For separable metric spaces (like $\Bbb R^n$ and most manifolds, like spheres etc.)  these notions all give the same values and agree that the value on $\Bbb R^n$ is $n$ (but this is quite hard to prove, mind you).
So in some sense it's true (by definition of dimension, plus some theorems), but only for boundaries of open sets. 
A: The boundary of $\mathbb Q$ in $\mathbb R$ is $\mathbb R$!
A: For a counterexample take the Cantor set. It, and every nonempty subset of it, is $0$-dimensional. It does have a certain nonempty subsets whose boundary is empty, i.e. $-1$-dimensional, for example the clopen subsets. Nonetheless many subsets have nonempty boundary, which gives lots of $0$-dimensional subsets with $0$-dimensional boundary; and this includes many open subsets.
A: You are mixing the notions of "a topological space" and "a subspace of a topological space".
The line, homoemorphic to $\Bbb{R}$, taken as a topological space, has no boundary.  It does not float in some enclosing space.  Open balls in the line are open intervals.
However, if we think of the line as a subspace of the plane, inheriting the subspace topology (by intersections of the line with open set of the plane), we find that every point of the line-in-the-plane is a boundary point.
The difference is the existence of the ambient space and whose open sets we are taking to define our topology.
