Linear maps from $\mathbb{R}^n$ to $\mathbb{R}^n$ are continuous I would like to prove that all linear maps $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ are continuous.
The following definition of continuity is the one I need to use: $f$ is continuous if the inverse image of any open subset of $\mathbb{R}^n$ is open in $\mathbb{R}^n$. I know that being linear means that $f(x+y)$ and $f(ax)=af(x)$ for $x,y\in \mathbb{R}^n$ and $a\in \mathbb{R}$, but how can I use these to prove the above?
 A: With respect to the standard basis $e^1,...,e^n$ on $\mathbb{R}^n$ you can express any linear map $f$ in the form;
$$\\$$
$$f(\textbf{x}) = \begin{bmatrix} T(e^1) & T(e^2) & \cdots & T(e^n) \end{bmatrix} \begin{bmatrix} x^1 \\ x^2 \\ \vdots \\ x^n \end{bmatrix}$$
$$\\$$
The result of the matrix multiplication is just a polynomial. This is just a special case, but you can see that with respect to any basis you place on $\mathbb{R}^n$, a linear map between finite-dimension vector spaces has the form $L(x) = Ax$ where $A$ is a matrix i.e $L$ is a polynomial; hence continuous. 
A: Hint 1: Since $f(x+y) =f(x)+f(y)$ it is enough to check continuity at $y=0$.
Hint 2: Since $f(ax)=af(x)$ it is enough to check that the preimage of the unit ball centered at $0$ is open.
Hint 3: Pick $m= \max\{ \|f(e_j)\| | 1 \leq j \leq n \}$ where $e_j$ is the standard basis. 
Now, if $m=0$ you are done. Otherwise, show that the pre-image of the unit ball centered at $0$ contains a ball of radius $\frac{1}{m}$  centered at $0$. 
A: First a note about linear functions:
$$d(f(ax),f(ay))=d(a f(x), a f(y)) = a \cdot d(x,y)$$
This will come in handy soon. Let us fix $a \in \mathbb R^n$.

Let $U \subseteq \mathbb R^n$ be open. Then $U$ consists entirely of interior points. i.e. for all $\mathbf x$ in $U$, there is an open neighbourhood $U_{\mathbf x}$ of $\mathbf x$ residing in $U$. Let us take an open ball of radius $R$ around $\mathbf x$ to be this $U_{\mathbf x}$.
$$U_{\mathbf x} = \{\mathbf y : d(\mathbf y, \mathbf x) < R\} \; .$$
Let us now suppose that the preimage of $U_{\mathbf x}$, call it $V_{\mathbf x}$, contains a point that isn't interior. Call this point $\mathbf p$. When mapped into $U_{\mathbf x}$, $\mathbf p$ is a certain distance from $x$ less than $R$. This means that
$$R - d(f(\mathbf p), \mathbf x) := r$$
is positive. Well, because there's some wiggle room around where $\mathbf p$ is mapped to, there should be some wiggle room about $\mathbf p$ as well-- i.e. it's an interior point. You could complete this proof with a diagram (recommended) but here's the rest of it in words:
Explicitly construct the set:
$$B_{\mathbf p} = \{\mathbf b : d(\mathbf b - \mathbf p) < r/a\}$$
It will be shown that this set resides in the preimage of $U_{\mathbf x}$, showing that $\mathbf p$ is in fact an interior point.
Each element of $B_{\mathbf p}$, when mapped via $f$, is less than $R$ away from $\mathbf x$ because:
$$\begin{align}
d(f(\mathbf b), \mathbf x) & \leq d(f(\mathbf b), f(\mathbf p)) + d(f(\mathbf p), \mathbf x) \\
&\leq a \cdot d(\mathbf b, \mathbf p) + (R-r) \\
&\leq a \cdot (r/a) + (R - r) \\
&\leq R
\end{align}$$
What this shows is that every point in the preimage of $U_{\mathbf x}$ is itself an interior point of $f^{-1}(U_{\mathbf x})$-- i.e. the preimage of the open set $U_{\mathbf x}$ is open.
Now $U$ is equal to the union of these open neighbourhoods for every point inside:
$$U = \bigcup_{\mathbf x \in U} U_{\mathbf x}$$
So that its preimage is the union of open sets:
$$f^{-1}(U) = \bigcup_{\mathbf x \in U} f^{-1}(U_{\mathbf x})$$
and so is itself open.
A: For $x= \left(x_1,\dots, x_n\right)' \in \mathbb{R}^n$, we know $x=\sum_{i=1}^n x_i e_i$ where $x_i\in \mathbb{R}$ is the ith entry of x and column vector $e_i$ is 0 except for its ith entry which is 1.  Use the linearity of the mapping, we the would have
$$f(x) =f(\sum_{i=1}^n x_i e_i) = \sum_{i=1}^n x_i f(e_i) = \left(f(e_1),\dots, f(e_n) \right) \left(x_1,\dots, x_n\right)' = Ax $$
Then we know $|f(x)-f(x')|=|A(x-x')|$ and continuity follows this clearly.
