$T:X\to Y$ a linear operator, $Y$ finite-dimensional, then $\ker T$ is closed iff $T$ is continuous I have to prove that if $T:X\to Y$ a linear operator on normed spaces, $Y$ finite-dimensional, then $\ker T$ is closed iff $T$ is continuous.

On a lot of places I see a proof that looks like this: it starts by assuming $Y=\mathbb{R}$ for simplicity. (I'm also aware of this that use a completely different construction). Anyway, I did not at all find it obvious that there was any kind of generalisation of the first linked proof to higher dimensions.

I will give my (quite long a tedious) generalisation here, but my question is: is there a more direct/easier generalisation of the construction given in the first linked proof? (I also wouldn't object to someone checking the actual correctness of this proof)



We will take $Y=\mathbb{R}^n$.

Take $x_n\in X$ s.t. $||x_n||=1$, $||Tx_n||\to \infty$. w.l.o.g. we may assume that $\frac{Tx_n}{||Tx_n||}\to v_1\in \overline{B(0,1)}\subset \mathbb{R^n}$, by the compactness of the closed unit ball in finite dimensions. We now take $y_n=x_j$ for some $j$ s.t. $||\frac{Tx_n}{||Tx_n||}-v_1||&lt;\frac{1}{n}$. Then we adjust the $y_n$ a little bit: if $||Ty_n||&gt;n$ we set $y_n=\frac{y_n}{||Ty_n||}n$. We now obtain a sequence $y_n$ s.t.
$$
\begin{align*}
||y_n||\leq 1\\
||Ty_n||\leq n\\
||Ty_n||\to \infty
\end{align*}
$$
In particular we have that $||Ty_n-||Ty_n||v_1||&lt;1$ (this is the critical property that we needed).

We now take an orthonormal basis $\{v_i\}$ for $\mathcal{R}T$ containing $v_1$, and $v_i'$ s.t. $Tv_i'=v_i$. We conclude that for $j&gt;1$ we must have $(v_j,Ty_n)&lt;1$. We now define $z_n=y_n-\sum_{j&gt;1}(v_j,Ty_n)v_j'$. Then
$$||z_n||=||y_n-\sum_{j&gt;1}(v_j,Ty_n)v_j'||\leq ||y_n||+\sum_{j&gt;1}(v_j,Ty_n)||v_j'||\leq 1+\sum_{j&gt;1}||v_j'||\leq C$$
Hence these $z_n$ are bounded. Furthermore we clearly still have that
$$||Tz_n||\to\infty$$
because the norm of $T\sum_{j&gt;1}(v_j,Ty_n)v_j'$ is bounded. 

We can now finally define 
$$w_n=v'_1-\frac{z_n}{||Tz_n||}$$
Then 
$$Tw_n=v_1-\frac{1}{||Tz_n||}(Ty_n-\sum_{j&gt;1}(v_j,Ty_n)v_j)=0$$
Since $Ty_n-\sum_{j&gt;1}(v_j,Ty_n)v_j$ is by construction a vector pointing in the direction of $v_1$. But again $w_n\to v'_1$, which contradicts the closedness of the kernel.



For the intuition: We take the unbounded sequence in $X$. The the project is onto the unit ball, which is compact, hence the sequence a convergent subsequence. We take the sequence, with limit $v_1$. Then we 'limit the growth' of our unbounded sequence to make sure that if $\frac{Tx}{||Tx||}$ is close to $v_1$, then $Tx$ is close to the line spanned by $v_1$. Then we remove any orthogonal components from the $Tx$, and because $Tx$ is now close to $&lt;v_1&gt;$ we can remove the orthogonal components by subtracting a vector of a certain bounded length. This finally gives a bounded sequence in $X$ with unbounded images in $Y$ all lying on the lines $&lt;v_1&gt;$, and from here the previous argument goes through.

Thanks for reading.
If $T$ is continuous, then $\ker(T) = T^{-1}(\{0\})$ is clearly closed. Conversely, suppose $Z:= \ker(T)$ is closed, then the quotient map
$$
\pi : X\to X/Z
$$
is a well-defined continuous map between normed linear spaces. The first isomorphism theorem (from linear algebra) gives an injective map
$$
S : X/Z \to Y
$$
such that $T = S\circ \pi$. Now $S$ is an injective map, so $X/Z$ must be finite dimensional, and hence $S$ is continuous, and so it follows that $T$ must be continuous as well.
If $T$ is continuous, then $\ker(T) = T^{-1}(\{0\})$ is clearly closed. Conversely, suppose $Z:= \ker(T)$ is closed, then the quotient map
$$
\pi : X\to X/Z
$$
is a well-defined continuous map between normed linear spaces. The first isomorphism theorem (from linear algebra) gives an injective map
$$
S : X/Z \to Y
$$
such that $T = S\circ \pi$. Now $S$ is an injective map, so $X/Z$ must be finite dimensional, and hence $S$ is continuous, and so it follows that $T$ must be continuous as well.