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 \( Y \)