Continuous Functions

Continuous Function

Let $ X, Y $ be topological spaces, a function $ f: X \rightarrow Y $ is said to be continuous iff for each open subset $ V $ of $ Y $, $ f ^{ -1 } \left( V \right) $ is an open set of $ X $

Continuous iff Every Basis Element is Open

Suppose that $ f : X \rightarrow Y $ is a function and $ \mathcal{ T }_Y $ is generated by a basis $ \mathcal{ B } $, then $ f $ is continuous if every basis element $ f ^ { -1 } \left( B \right) $ is open in $ Y $

Let $ U $ be open in $ Y $, we want to prove that $ f ^ { -1 } \left( U \right) $ is an open set. Since $ U $ is open then we know that $ U = \bigcup \mathcal{ C } $ where $ \mathcal{ C } \subseteq \mathcal{ B } $.

By noting that $ f ^ { -1 } \left( U \right) = f ^ { -1 } \left( \bigcup \mathcal{ C } \right) $ $ = $ $ \bigcup \left\{ f ^ { -1 } \left( C \right) : C \in \mathcal{ C } \right\} $, since $ \mathcal{ C } \subseteq \mathcal{ B } $, then every element $ C $ is a basis element, and thus $ f ^ { -1 } \left( C \right) $ is an open set by our assumption, therefore as it is an arbitrary union of open sets, we can see that $ f ^ { -1 } \left( U \right) $ is an open set, as needed

Continuous iff Every Subbasis Element is Open

Suppose that $ f : X \rightarrow Y $ is a function and $ \mathcal{ T }_Y $ is generated by a subbasis $ \mathcal{ S } $, then $ f $ is continuous if every subbasis element $ f ^ { -1 } \left( S \right) $ is open in $ X $

Since $ \mathcal{ T }_Y $ is generated by $ S $, this means that $ \mathcal{ T }_Y = \mathcal{ T }_{ \mathcal{B}_S } $ where $ \mathcal{ B }_S $ is the basis generated by $ S $ therefore to verify that this function is continuous all we have to do is check that every basis element $ B \in \mathcal{ B }_S $ has the property that $ f ^ { -1 } \left( B \right) $ is open with respect to $ X $

We know $ B = S_1 \cap S_2 \cap \ldots \cap S_k $ for some $ k \in \mathbb{ N }_1 $, therefore we want to know if $ f ^ { -1 } \left( S_1 \cap S_2 \cap \ldots \cap S_k \right) $ is open, since we know that the intersection factors through the inverse image then we can re-write this as $ \bigcap _ { i = 1 } ^ { k } f ^ { -1 } \left( S_i \right) $.

Our initial assumption was that given any $ S \in \mathcal{ S } $ that $ f ^ { -1 } \left( S \right) $ was open, thus specifically for our $ S_1, \ldots , S_k $ their inverse images should also be open, therefore $ \bigcap _ { i = 1 } ^ { k } f ^ { -1 } \left( S_i \right) $ is a finite intersection of open sets of $ X $ thus it is open, which shows that $ f $ is continuous.

Continuous Implies Image Closure Expansion

Suppose that $ f $ is continuous, then given any subset $ A $ of $ X $, we have that $ f \left( \overline{ A } \right) \subseteq \overline{ f \left( A \right) } $

Suppose that $ f $ is continuous, and suppose that $ A \subseteq X $, we want to show that $ f \left( \overline{ A } \right) \subseteq \overline{ f \left( A \right) } $, we start by assuming that $ p \in f \left( \overline{ A } \right) $, which means that $ p = f \left( a \right) $ where $ a \in \overline{ A } $, we now want to prove that $ p \in \overline{ f \left( A \right) } $, one way of going about this would be to show that every neighborhood of $ p $ intersects $ f \left( A \right) $, which we will do.

So let $ V $ be a neighborhood of $ p $, by definition we know $ V \subseteq Y $, then we know that $ f ^ { -1 } \left( V \right) $ is an open set since $ f $ is continuous, since $p = f \left( a \right) $, then $ V $ is also a neighborhood of $ f \left( a \right) $, therefore clearly $ f \left( a \right) \in V $, that is $ a \in f ^ { -1 } \left( V \right) $.

We assumed that $ a \in \overline{ A } $, which is equivalent to every neighborhood of $ a $ intersecting $ A $, we know that $ f ^ { -1 } \left( V \right) $ is an open set containing $ a $, which makes it a neighborhood of $ a $, and thus it must intersect with $ A $ at some point $ b $, then $ f \left( b \right) \in V $, at the same time we know that $ b \in A $ as well, so that $ f \left( b \right) \in f \left( A \right) $, thus $ f \left( b \right) \in V \cap f \left( A \right) $ showing that the neighborhood intersects $ f \left( A \right) $, therefore this shows that $ p \in \overline{ f \left( A \right) } $, as needed.

Closure Expansion Implies Closed Inverse Image

Suppose that for every subset $ A \subseteq X $, we have $ f \left( \overline{ A } \right) \subseteq \overline{ f \left( A \right) } $, then for every closed set $ B $ of $ Y $, $ f ^ { -1 } \left( B \right) $ is closed in $ X $

Let $ B \subseteq Y $ be a closed set of $ Y $, we want to prove that $ A = f ^ { -1 } \left( B \right) $ is a closed set of $ X $, which is equivalent to $ A = \overline{ A } $, therefore we will prove this instead.

We already know $ A \subseteq \overline{ A } $, so all we need to show is that $ \overline{ A } \subseteq A $, so let $ x \in \overline{ A } $, and we want to prove that $ x \in A = f ^ { -1 } \left( B \right) $, in other words we want to show that $ f \left( x \right) \in B $.

Before we continue, we should realize that $ A $ is a subset of $ X $, therefore our assumption applies, which tells us $ f \left( \overline{ A } \right) = \overline{ f \left( A \right) } $, since $ A = f ^ { -1 } \left( B \right) $, we can conclude that $ f \left( \overline{ A } \right) \subseteq B $.

Since $ x \in \overline{ A } $, then $ f \left( x \right) \in f \left( \overline{ A } \right) $, but then by our previous paragraph, we know that $ f \left( x \right) \in B $, as needed.

Closed Inverse Image Implies Continuous

Suppose that for every closed $ B $ in $ Y $, $ f ^ { -1 } \left( B \right) $ is closed in $ X $, then $ f $ is continuous

Let $ V $ be an open set of $ Y $, we want to show that $ f ^ { -1 } \left( V \right) $ is an open set of $ X $, set $ B = Y \setminus V $.

Since $ V $ was open, then $ B $ is closed, and thus $ f ^ { -1 } \left( B \right) $ is a closed set. But note this: \[ \begin{align} f ^ { -1 } \left( B \right) &= f ^ { -1 } \left( Y \setminus V \right) \\ &= f ^ { -1 } \left( Y \right) \setminus f ^ { -1 } \left( V \right) \tag{$\alpha$} \\ &= X \setminus f ^ { -1 } \left( V \right) \tag{$\beta$} \end{align} \]

Where we've used the facts:

The conclusion from the above was that $ f ^ { -1 } \left( B \right) = X \setminus f ^ { -1 } \left( V \right) $, from our assumption we knew that $ f ^ { -1 } \left( V \right) $ was a closed, set, which makes $ f ^ { -1 } \left( B \right) $ an open set, which is what we set out to prove.

Continuous iff Every Neighborhood Contains the Image of another Neighborhood

$ f $ is continuous iff for every $ x \in X$ and neighborhood $ V $ of $ f \left( x \right) $, there is a neighborhood $ U $ of $ x $ such that $ f \left( U \right) \subseteq V $

$ \implies $ Let $ x \in X $ and suppose $ V $ is a neighborhood of $ f \left( x \right) $, then $ f ^ { -1 } \left( V \right) $ is an open set since $ f $ was assumed continuous, it also must contain $ x $, this is because $ f \left( x \right) \in V $, making $ f ^ { -1 } \left( V \right) $ a neighborhood of $ x $, we also know $ f \left( f ^ { -1 } \left( V \right) \right) \subseteq V $, as needed.

$ \impliedby $ Suppose $ V $ is open in $ Y $, if $ f ^ { -1 } \left( V \right) = \emptyset $, we are done, thus we can find some element $ x \in f ^ { -1 } \left( V \right) $, making $ V $ a neighborhood of $ f \left( x \right) $, thus by assumption, we get some neighborhood $ U _ x $ such that $ f \left( U _ x \right) \subseteq V $, this by definition means that for any $ p \in U _ x $, we know $ f \left( p \right) \in V $, or that $ p \in f ^ { -1 } \left( V \right) $, which shows that $ U _ x \subseteq f ^ { -1 } \left( V \right) $.

Since this is true for every point $ x \in f ^ { -1 } \left( V \right) $, since $ \bigcup _ { x \in f ^ { -1 } \left( V \right) } U _ x = f ^ { -1 } \left( V \right) $, then $ f ^ { -1 } \left( V \right) $ can be written as an arbitrary union of open sets, and thus it is open, as needed.

Continuous, Image Closure Expansion, Closed Inverse Image, Nested Neighborhoods Equivalence

Let $ X, Y $ be topological spaces, let $ f: X \rightarrow Y $, then the following are equivalent:

$ f $ is continuous
For every subset $ A $ of $ X $, $ f \left( \overline{ A } \right) \subseteq \overline{ f \left( A \right) } $
For every closed set $ B $ of $ Y $, $ f ^ { -1 } \left( B \right) $ is closed in $ X $
For each $ x \in X $ and each neighborhood $ V $ of $ f \left( x \right) $, there is a neighborhood $ U $ of $ x $ such that $ f \left( U \right) \subseteq V $

We've shown that $ 1 \implies 2 $ , $ 2 \implies 3 $, $ 3 \implies 4 $ and $ 4 \iff 1 $, this induces directed graph that is weakly connected and has a cycle, therefore it is connected, which means that all statements are equivalent.

Homeomorphism

Let $ X, Y $ be topological spaces, and let $ f: X \rightarrow Y $ be a bijection. If the function $ f $ and it's inverse function $ f ^ { -1 } : Y \rightarrow X $ are continuous, then $ f $ is called a homeomorphism

Homeomorphism Set and it's Image are Open Equivalence

Suppose that $ f $ is a bijection, then $ f $ is a homeomorphism iff the following is true: $ f \left( U \right) $ is open in $ Y $ iff $ U $ is open in $ X $

Suppose that $ f $ is a homeomorphism, and that $ U $ is a subset of $ X $, assume $ f \left( U \right) $ is open in $ Y $ , we'll prove that $ U $ is open in $ X $, since $ f $ is continuous, then we know that $ f ^ { -1 } \left( f \left( U \right) \right) $ is an open set of $ X $, since $ f $ is a bijection, this means it's injective and thus $ f ^ { -1 } \left( f \left( U \right) \right) = U $ as needed.

Now suppose that $ U $ is open in $ X $, since we know that $ f ^ { -1 } $ is continuous this means that $ \left( f ^ { -1 } \right) ^ { -1 } \left( U \right) $ is an open set of $ Y $, we've shown that $ \left( f ^ { -1 } \right) ^ { -1 } \left( U \right) = f \left( U \right) $, thus, we've just shown that $ f \left( U \right) $ is an open set of $ Y $, this concludes

At this point we've proven the main rightward implication. The other direction is quite a lot easier. We have to show that both $ f $ and $ f ^ { -1 } $ are continuous.

Suppose that $ V $ is an open set of $ Y $, we want to show that $ f ^ { -1 } \left( V \right) $ is an open set of $ X $, which is true iff $ f \left( f ^ { -1 } \left( V \right) \right) $ is an open set of $ Y $, since $ f $ is surjective, then $ f \left( f ^ { -1 } \left( V \right) \right) = V $ making it an open set of $ Y $, which shows that $ f ^ { -1 } \left( V \right) $ is an open set of $ X $

Now suppose that $ U $ is an open set of $ X $ , our goal will be to show that $ \left( f ^ { -1 } \right) ^ { -1 } \left( U \right) $ is an open set of $ Y $, as seen previously $ \left( f ^ { -1 } \right) ^ { -1 } \left( U \right) = f \left( U \right) $, and since $ U $ was open in $ X $, by our assmption we know that $ f \left( U \right) $ is open in $ Y $, this concludes the proof.

Imbedding

Suppose that $ f: X \rightarrow Y $ is an injective continuous function, with $ X $ and $ Y $ topological spaces. By setting $ Z = f \left( X \right) $ as a subspace of $ Y $, if $ f ^ \prime : X \rightarrow Z $ obtained by restricting the range of $ f $ is a homeomorphism, then we say that the original $ f $ is an imbedding of $ X $ in $ Y $

Constant Functions are Continuous

Let $ X, Y $ be topological spaces and suppose that $ p \in Y $, if $ f: X \rightarrow Y $ is constant with value $ p $, then $ f $ is continuous

Suppose that $ V \subseteq Y $, if $ p \in V $, then $ f ^ { -1 } \left( V \right) = X $, otherwise $ p \notin V $ so $ f ^ { -1 } \left( V \right) = \emptyset $, in either case we have an open set of $ X $.

Inclusion is Continuous

Suppose that $ A $ is a subspace of $ X $, then the inclusion function $ \iota: A \rightarrow X $ is continuous.

Let $ U $ be an open set of $ X $, then $ \iota ^ { -1 } \left( U \right) = U \cap A $, which by definition is open in the subspace topology of $ A $ as needed.

Compositions of Continuous Functions are Continuous

Suppose that $ f : X \rightarrow Y $ and $ g: Y \rightarrow Z $ are continuous then the map $ g \circ f : X \to Z $ is continuous

Let $ U $ be open in $ Z $, we want to prove that $ \left( g \circ f \right) ^ { -1 } \left( U \right) $ is an open set of $ X $.

To start if we look at $ \left( g \circ f \right) ^ { -1 } \left( U \right) $, we know this is equal to $ f ^ { -1 } \left( g ^ { -1 } \left( U \right) \right) $, thus since $ g $ is continuous we know that $ S := g ^ { -1 } \left( U \right) $ is an open set of $ Y $ and then $ f ^ { -1 } \left( S \right) $ is open in $ X $ since $ f $ was continuous, this shows that $ g \circ f $ is continuous

Domain Restriction is Continuous

If $ f : X \to Y $ is continuous and if $ A $ is a subspace of $ X $, then the restricted function $ f \restriction A : A \to Y $ is continuous

We note that $ f \restriction A = f \circ \iota $, therefore as the composition of two continuous functions, we know that $ f \restriction A $ is continuous

Restricting the Range maintains Continuity

Suppose that $ f : X \to Y $ is continuous. If $ Z $ is a subspace of $ Y $ containing $ f \left( X \right) $, then the function $ g: X \to Z $ obtained by restricting the range is continuous

To show continuity we start by assuming that $ B $ is open in $ Z $, therefore $ B = Z \cap U $ where $ U $ is open in $ Y $