> [!theoremb] Theorem > > Let $\cx$ and $\cy$ be [[Banach Space|Banach spaces]], and $T \in L(\cx, \cy)$ be a [[Bounded Linear Map|continuous linear map]]. If $T$ is surjective, then $T$ is [[Open Map|open]]. > > *Proof*. Using the [[Baire Category Theorem]], first find an open ball in the *closure* of $T(B(0, 1))$. Translate the ball to $0$ yields a ball centred at $0$ in the closure. Since elements of the closure can be approximated with elements of the ball, the [[Method of Successive Approximations]] can be applied to construct the precise preimage of each element in the ball. > > # Finding the Open Ball > > Let $\cx$ be a [[Normed Vector Space|normed space]], $\cy$ be a Banach space, and $T:\cx \to \cy$ be a linear map. If $T$ is surjective, then there exists a non-empty open set $U \subset \ol{T(B(0, 1))}$. > > *Proof*. Let $B_n = B(0, n)$, then $\cx = \bigcup_{n \in \nat}B_n$. As $T$ is surjective, $\cy = \bigcup_{n \in \nat}T(B_n) = \bigcup_{n \in \nat}\ol{T(B_n)}$. With $T$ being linear, > $ > \ol{T(B_n)} = \ol{T(nB_1)} = \ol{nT(B_1)} = n\ol{T(B_1)} > $ > Since $\cy$ is a Banach space, by the [[Baire Category Theorem]], there exists $n \in \nat$ such that $(\ol{T(B_n)})^o \ne \emptyset$. Undoing the scaling[^1], > $ > \paren{\ol{T(B_1)}}^o = \frac{1}{n}\paren{\ol{nT(B_1)}}^o = \frac{1}{n}\paren{\ol{T(B_n)}}^o \ne \emptyset > $ > and we have found a non-empty open set. > > > # Moving the Ball > > Not being nowhere dense means being dense somewhere, in this case the ball. Since this ball is an image by a linear operator, moving its "preimage" allows moving the ball itself. > > Let $\cx$ and $\cy$ be normed spaces and $T: \cx \to \cy$ be a linear map. If $\ol{T(B(0, 1))}$ has a non-empty interior, then it contains an open ball centred at $0$. > > *Proof*. Let $y_0 \in \ol{T(B_{1/2})}$, then since $(\ol{T(B_{1/2})})^o = \frac{1}{2}(\ol{T(B_{1})})^o$ is non-empty, there exists $\varepsilon > 0$ such that $B(y_0, \varepsilon) \subset \ol{T(B_{1/2})}$. > > Using the fact that $T(B_{1/2})$ is dense in $\ol{T(B_{1/2})}$, choose $x_1 \in B_{1/2}$ such that $\norm{y - Tx_1} < \varepsilon/2$, then > $ > B(Tx_1, \varepsilon/2) \subset B(y_0, \varepsilon) \subset \ol{T(B_{1/2})} > $ > and we have created a ball centred at $y_1 = Tx_1 \in T(B_{1/2})$, which allows us to move another open ball there while maintaining its connection with $B_{1}$. > > To see this, let $y \in B(0, \varepsilon / 2)$, then > $ > \begin{align*} > y &= y + y_1 - y_1 = \underbrace{y + y_1}_{\in \ol{T(B_{1/2})}} - \underbrace{Tx_1}_{\in T(B_{1/2})} \in \ol{T(B_1)} > \end{align*} > $ > and we have found an open ball $B(0, \varepsilon/2) \subset \ol{T(B_1))}$. > > > # Successive Approximations > > Lastly, we can approximate any element in $\ol{T(B_1)}$ closely using elements of $T(B_1)$ simply because it is dense in its closure. > > Let $\cx$ and $\cy$ be Banach spaces and $T \in L(\cx, \cy)$ be a [[Bounded Linear Functional|continuous linear functional]]. If there exists $\varepsilon > 0$ such that $B(0, \varepsilon) \subset \ol{T(B_1)}$, then $B(0, \varepsilon) \subset T(B_1)$. > > *Proof, with [[Method of Successive Approximations]] on $\Sigma = T(B_1)$*. > > Let $r \in (0, 1]$, then $B(0, r\varepsilon) \subset \ol{T(B_r)}$. Let $y \in B(0, r\varepsilon) \subset B(0, \varepsilon)$, then since $T(B_r)$ is dense in $\ol{T(B_r)} \supset B(0, r\varepsilon)$, there exists $Tx \in T(B_r) \subset T(B_1)$ such that $\norm{y - Tx} < \frac{1}{2}r\varepsilon$ and $\norm{x} < r$. > > Since $\cx, \cy$ is complete and $T \in L(\cx, \cy)$ is continuous, there exists a sequence $\seq{x_n} \subset B_1$ such that > $ > T\paren{\sum_{n =1}^{\infty}x_n} = y > $ > for any $y \in \ol{T(B_1)}$. > > > # Combining the Arguments > > Let $\cx$ and $\cy$ be Banach spaces, and $T \in L(\cx, \cy)$ be a continuous linear map. If $T$ is surjective, then $T$ is open. > > *Proof*. Since $\cy$ is a Banach space and $T$ is surjective, there exists an open ball $B(y, \varepsilon) \subset \ol{T(B_1)}$. This means that there exists an open ball $B(0, \varepsilon') \subset \ol{T(B_1)}$ centred at $0$. Since $\cx$ is a Banach space, we can use successive approximations to approach any element in $B(0, \varepsilon')$ using elements of $T(B_1)$, removing the need for a closure using completeness. > > Therefore $B(0, \varepsilon') \subset T(B_1)$ and $T$ is an open map. > [!theorem] > > Let $\cx, \cy$ be Banach spaces and $T \in L(\cx, \cy)$ be bijective, then $T$ is an isomorphism with $T^{-1} \in L(\cy, \cx)$. > > *Proof*. $T \in L(\cx, \cy)$ is surjective and open, $T^{-1}(\cy, \cx)$ is surjective and open. For any open sets $U \subset \cx$, $V \subset \cy$, > $ > T^{-1}(V) = \paren{T^{-1}}(V) \quad \paren{T^{-1}}^{-1}(U) = T(U) > $ > are both open. Therefore $T^{-1} \in L(\cy, \cx)$ both ways. [^1]: [[Translation and Dilation of Closure and Interior]]