Mapping is Continuous iff Inverse Images of Open Sets are Open/Corollary
Theorem
Let $X$ and $Y$ be normed vector spaces.
Let $f : X \to Y$ be a mapping.
Then:
- $f$ is continuous on $X$
Proof
Sufficient Condition
Suppose that for every closed $F$ in $Y$, $\map {f^{-1}} F$ is closed in $X$.
Let $V$ be open in $Y$.
By definition, $Y \setminus V$ is closed in $Y$.
By assumption, $\map {f^{-1}} {Y \setminus V}$ is closed in $X$.
We have that:
| \(\ds \map {f^{-1} } {Y \setminus V}\) | \(=\) | \(\ds \map {f^{-1} } Y \setminus \map {f^{-1} } V\) | Preimage of Set Difference under Mapping | |||||||||||
| \(\ds \) | \(=\) | \(\ds X \setminus \map {f^{-1} } V\) | Definition of $f$ |
Thus, $X \setminus \map {f^{-1} } V$ is closed in $X$.
By definition, $X \setminus \map {f^{-1} } V$ is closed iff $\map {f^{-1} } V$ is open.
Hence, for every $V$ open in $Y$, $ \map {f^{-1}} V$ is open in $X$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $f$ is continuous on $X$.
$\Box$
Necessary Condition
Suppose that $f$ is continuous on $X$.
Let $F$ be closed in $Y$.
By definition, $Y \setminus F$ is open in $Y$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $\map {f^{-1} } {Y \setminus F}$ is open on $X$.
Hence:
| \(\ds \map {f^{-1} } {Y \setminus F}\) | \(=\) | \(\ds \map {f^{-1} } Y \setminus \map {f^{-1} } F\) | Preimage of Set Difference under Mapping | |||||||||||
| \(\ds \) | \(=\) | \(\ds X \setminus \map {f^{-1} } F\) | Definition of $f$ |
and $X \setminus \map {f^{-1} } F$ is open in $X$.
By definition, $\map {f^{-1}} F$ is closed in $X$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.2$: Continuous and linear maps. Continuous maps