## What is open mapping

An open map is a function between two topological spaces that maps open sets to open sets. Precisely speaking, a function $f: X \to Y$ is open if for any open set $U \subset X$, $f(U)$ is open in $Y$. Likewise, a closed map is a function mapping closed sets to closed sets.

You may think open/closed map is an alternative name of continuous function. But it’s not. The definition of open/closed mapping is totally different from continuity. Here are some simple examples.

- $f(x)=\sin{x}$ defined on $\mathbb{R}$ is not open, though it’s continuous. It can be verified by considering $(0,2\pi)$, since we have $f((0,2\pi))=[-1,1]$.
- The projection $\pi: \mathbb{R}^2 \to \mathbb{R}$ defined by $(x,y) \mapsto x$ is open. Indeed, it maps an open ball onto an open interval on $x$ axis.
- The inclusion map $\varphi: \mathbb{R} \to \mathbb{R}^2$ by $x \mapsto (x,0)$ however, is not open. An open interval on the plane is
*locally closed*but not open or closed.

Under what condition will a continuous linear function between two TVS be an open mapping? We’ll give the answer in this blog post. Open mapping theorem is a sufficient condition on whether a continuous linear function is open.

## Open Mapping Theorem

Let $X,Y$ be Banach spaces and $T: X \to Y$ a

surjectivebounded linear map. Then $T$ is an open mapping.

The open balls in $X$ and $Y$ are defined respectively by

All we need to do is show that there exists some $r>0$ such that

Since every open set in $X$ or $Y$ can be expressed as a union of open balls. For a ball in $X$ centered at $x \in X$ with radius $r$, we can express it as $x+B_r^X$. After that, it becomes obvious that $T$ maps open set to open set.

First we have

The surjectivity of $T$ ensures that

Since $Y$ is Banach, or simply a complete metric space, by Baire category theorem, there must be some $n_0 \in \mathbb{N}$ such that $\overline{T(B_{n_0}^{X})}$ has nonempty interior. If not, which means $T(B_n^{X})$ is nowhere dense for all $n \in \mathbb{N}$, we have $Y$ is of the first category. A contradiction.

Since $x \to nx$ is a homeomorphism of $X$ onto $X$, we see in fact $T(B_n^X)$ is not nowhere dense for all $n \in \mathbb{N}$. Therefore, there exists some $y_0 \in \overline{T(B_1^{X})}$ and some $\varepsilon>0$ such that

the open set on the left hand is a neighborhood of $y_0$, which should be in the interior of $\overline{T(B_1^X)}$.

On the other hand, we claim

We shall prove it as follows. Pick any $y \in \overline{T(B_1^X)}$, we shall show that $y-y_0 \in \overline{T(B_2^X)}$. For $y_0$, there exists a sequence of $y_n$ where $\lVert y_n \rVert <1$ for all $n$ such that $Ty_n \to y_0$. Also we are able to find a sequence of $x_n$ where $\lVert x_n \rVert <1$ for all $n$ such that $Tx_n \to y$. Notice that we also have

since

we see $T(x_n-y_n) \in T(B_2^X)$ for all $n$, it follows that

Combining all these relations, we get

Since $T$ is linear, we see

By induction we get

for all $n \geq 1$.

We shall show however

For any $u \in B_{\varepsilon/4}^Y$, we have $u \in \overline{T(B_{1/2}^X)}$. There exists some $x_1 \in B_{1/2}^{X}$ such that

This implies that $u-Tx_1 \in B_{\varepsilon/8}^Y$. Under the same fashion, we are able to pick $x_n$ in such a way that

where $\lVert x_n \rVert<2^{-n}$. Now let $z_n=\sum_{k=1}^{n}x_k$, we shall show that $(z_n)$ is Cauchy. For $m<n$, we have

Since $X$ is Banach, there exists some $z \in X$ such that $z_n \to z$. Further we have

therefore $z \in B_1^X$. Since $T$ is bounded, therefore continuous, we get $T(z)=u$. To summarize, for $u \in B_{\varepsilon/4}^Y$, we have some $z \in B_{1}^X$ such that $T(z)=y$, which implies $T(B_1^X) \supset B_{\varepsilon/4}^Y$.

Let $U \subset X$ be open, we want to show that $T(U)$ is also open. Take $y \in T(U)$, then $y=T(x)$ with $x \in U$. Since $U$ is open, there exists some $\varepsilon>0$ such that $B_{\varepsilon}^{X}+x \subset U$. By the linearity of $T$, we obtain $B_{r\varepsilon}^Y \subset T(B_{\varepsilon}^X)$ for some small $r$. Using the linearity of $T$ again, we obtain

which shows that $T(U)$ is open, therefore $T$ is an open mapping.

### Remarks

One have to notice that the completeness of $X$ and $Y$ has been used more than one time. For example, the existence of $z$ depends on the fact that Cauchy sequence converges in $X$. Also, the surjectivity of $T$ cannot be omitted, can you see why?

There are some different ways to state this theorem.

- To every $y$ with $\lVert y \rVert < \delta$, there corresponds an $x$ with $\lVert x \rVert<1$ such that $T(x)=y$.
- Let $U$ and $V$ be the open unit balls of the Banach spaces $X$ and $Y$. To every surjective bounded linear map, there corresponds a $\delta>0$ such that

You may also realize that we have used a lot of basic definitions of topology. For example, we checked the openness of $T(U)$ by using neighborhood. The set $\overline{T(B_1^X)}$ should also remind you of limit point.

The difference of open mapping and continuous mapping can be viewed via the topologies of two topological vector spaces. Suppose $f: X \to Y$. If for any $U \in \tau_X$, we have $f(U) \in \tau_Y$, where $\tau_X$ and $\tau_Y$ are the topologies of $X$ and $Y$, respectively. But this has nothing to do with continuity. By continuity we mean, for any $V \in \tau_Y$, we have $f^{-1}(V) \in \tau_U$.

Fortunately, this theorem can be generalized to $F$-spaces, which will be demonstrated in the following blog post of the series. A space $X$ is an $F$-space if its topology $\tau$ is induced by a complete invariant metric $d$. Still, completeness plays a critical rule.