Pasting Lemma/Union of Open Sets

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Theorem

Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.

Let $I$ be an indexing set.

Let $\family {C_i}_{i \mathop \in I}$ be a family of open sets of $T$.

Let $f: X \to Y$ be a mapping such that the restriction $f \restriction_{C_i}$ is continuous for all $i \in I$.


Then $f$ is continuous on $C = \ds \bigcup_{i \mathop \in I} C_i$, that is, $f \restriction_C$ is continuous.


Corollary 1

Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.

Let $I$ be an indexing set.

Let $\family {C_i}_{i \mathop \in I}$ be a family of open sets of $T$.

Let $\family {f_i : C_i \to Y}_{i \mathop \in I}$ be a family of continuous mappings.


Let $C = \ds \bigcup_{i \mathop \in I} C_i$.

Let $f = \ds \bigcup_{i \mathop \in I} f_i : C \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of relations.


Let for all $i, j \in I$, $f_i$ and $f_j$ agree on $C_i \cap C_j$.


Then $f$ is a continuous mapping on $C = \ds \bigcup_{i \mathop \in I} C_i$.


Corollary 2

Let $X$ and $Y$ be topological spaces.

Let $A$ and $B$ be open in $X$.

Let $f: A \to Y$ and $g: B \to Y$ be continuous mappings that agree on $A \cap B$.

Let $f \cup g$ be the union of the mappings $f$ and $g$:

$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$


Then the mapping $f \cup g : A \cup B \to Y$ is continuous.


Proof

Let $V$ be an open set of $S$.

By assumption, we have that, for all $i \in I$, $U_i = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is also open in $T$.

By Definition of Restriction of Mapping, we have that $U_i = C_i \cap f^{-1} \sqbrk V$.

Therefore, we can compute:

\(\ds \paren {f \restriction_C}^{-1} \sqbrk V\) \(=\) \(\ds C \cap f^{-1} \sqbrk V\) Definition of Restriction of Mapping
\(\ds \) \(=\) \(\ds \paren {\bigcup_{i \mathop \in I} C_i} \cap f^{-1} \sqbrk V\) Definition of $C$
\(\ds \) \(=\) \(\ds \bigcup_{i \mathop \in I} \paren {C_i \cap f^{-1} \sqbrk V}\) Intersection Distributes over Union
\(\ds \) \(=\) \(\ds \bigcup_{i \mathop \in I} U_i\) Definition of $U_i$

That is, $U = \paren {f \restriction_C}^{-1} \sqbrk V$ is a union of open sets.

Therefore, $U$ is itself open by definition of topology.

It follows that $f \restriction_C$ is also continuous by definition.

$\blacksquare$


Also see


Sources

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