Continuity

Section 2.2 Continuity

The good news now is that continuity works almost exactly as it did in the example of subspaces of Euclidean space (Section 1.1). The only difference is that we no langer have access to the

ε

δ

characterization of continuity.

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Definition 2.2.1.

Let $X$ and $Y$ be topological spaces, and let $f : X \to Y$ be a map. Then $f$ is continuous if and only if, for every subset $S \subseteq X$ any every point $x \in X$ that is close to $S,$ the point $f (x)$ is close to $f (S) .$

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Proposition 2.2.2.

Let $X$ and $Y$ be topological spaces. The following are equivalent for a map $f : X \to Y .$

The map $f$ is continuous.
For any subset $T \subseteq Y,$ one has $τ_{X} (f^{- 1} (T)) \subseteq f^{- 1} (τ_{Y} (T)) .$
For any closed subset $Z \subseteq Y,$ the inverse image $f^{- 1} (Z) \subseteq X$ is closed.
For any open subset $U \subseteq Y,$ the inverse image $f^{- 1} (U) \subseteq X$ is open.

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Proof.

Everything is exactly as in the proof of Proposition 1.2.2, except that since we do not have condition (5), we shall have to prove directly that (4) implies (1).

So assume (4); we aim to prove (1). Let $S \subseteq X\text{,}$ and let $x \in \tau_X(S)\text{.}$ Observe that the complement $V \coloneq Y \smallsetminus \tau_Y(f(S))$ is open; hence so is the inverse image $U \coloneq f^{-1}(V) \subseteq X\text{.}$ Note also that $S$ is disjoint from $U\text{.}$ Thus the complement $X \smallsetminus U$ is a closed subset that contains $S\text{.}$ Consequently, $\tau_X(S) \subseteq U\text{.}$ So since $x \in \tau_X(S)\text{,}$ it follows that $f(x) \notin V\text{.}$ In other words, $f(x) \in \tau_Y(f(S))\text{,}$ as desired.

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Example 2.2.3.

Let $X\text{,}$ $Y\text{,}$ and $Z$ be topological spaces. Then the identity map $\id \colon X \to X$ is continuous. Also, if $f \colon X \to Y$ and $g \colon Y \to Z$ are continuous, then the composition $g \circ f \colon X \to Z$ is continuous as well.

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Example 2.2.4.

Let $X$ be a topological space, and let $S$ be a set. Any map $f \colon S \to X$ is continuous if $S$ is endowed with the discrete topology. Dually, any map $g \colon X \to S$ is continuous if $S$ is endowed with the emph{chaotic} topology.

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Definition 2.2.5.

Suppose $τ_{1}$ and $τ_{2}$ are two topologies on the same set $X .$ Then we say that $τ_{1}$ is finer than $τ_{2}$ — and that $τ_{2}$ is coarser than $τ_{1}$ — if the identity map on $X$ is continuous as a map

(X, τ_{1}) (X, τ_{2}) .

Hence if $τ_{1}$ is finer than $τ_{2},$ then the closure of a set relative to $τ_{1}$ is contained in its closure relative to $τ_{2} .$ In particular, the topology $τ_{1}$ has more closed sets than the topology $τ_{2} .$ Forming complements, we deduce also that the topology $τ_{1}$ has more open sets than the topology $τ_{2} .$

There are two extremes: the finest topology on a set is the discrete topology, and the coarsest is the chaotic topology.

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Example 2.2.6.

Recall the topological space $\PP^1_{\RR}$ of Example 2.1.1. Define a map $L \colon S^1 \to \PP^1_{\RR}$ as follows: for every point $x \in S^1\text{,}$ let $L(x)$ be the $1$-dimensional subspace spanned by the vector $x\text{.}$ That is, $L(x)$ is the unique line in $\RR^2$ that passes through the origin and $x\text{.}$ It follows from the definition of the topology on $\PP^1_{\RR}$ that $L$ is continuous.

To unpack this a little, let $S \subseteq S^1$ be a subset, and let $x \in \tau(S)\text{.}$ We want to see that $L(x)$ is close to the image $L(S)\text{.}$ Let $\varepsilon>0\text{.}$ There exists $s \in S$ such that $\|x-s\| \lt \varepsilon\text{.}$ Now $L(x)$ intersects $S^1$ at the points $x$ and $-x\text{,}$ and $L(s)$ intersects $S^1$ in the points $s$ and $-s\text{.}$ Thus the line $L(s)$ has the property that, in the notation of Example 2.1.1, $\|i(L(x))-i(L(s))\| \lt \varepsilon$ or $\|i(L(x))-j(L(s))\| \lt \varepsilon\text{.}$

Please note that while $L$ is surjective, it is not injective, since for every $x \in S^1\text{,}$ one has $L(x) = L(-x)\text{.}$

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Definition 2.2.7.

Let $X$ and $Y$ be topological spaces, and let $x \in X .$ Then a map $f : X \to Y$ is continuous at $x$ if and only if, for any subset $S \subseteq X,$ if $x$ is close to $S$ then $f (x)$ is close to $f (S) .$

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A map

f : X \to Y

is continuous if and only if it is continuous at every point

x \in X .

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Example 2.2.8.

Consider the map $s \colon \RR \to \RR$ given by the formula

\begin{equation*} s(x) \coloneq \begin{cases}x/|x| \amp \text{ if } x \neq 0 ; \\ 0 \amp \text{ if } x=0 . \end{cases} \end{equation*}

Then $s$ is continuous at every $x \in \RR \smallsetminus \{0\}\text{.}$ If $s$ were continuous at $0\text{,}$ then it would be continuous. But even though the set $\{1\} \subset \RR$ is closed, its inverse image

\begin{equation*} s^{-1}\{1\} = \left]0, +\infty \right[ \subset \RR \end{equation*}

is not. Hence $s$ is not continuous at $0\text{,}$ as we expect!

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Example 2.2.9.

Consider the function $f(x)=1/x\text{.}$ This is a continuous map

\begin{equation*} f\colon \RR-\{0\} \to \RR-\{0\}\text{,} \end{equation*}

relative to the subspace topology on each side. Of course we have removed the point $0\in\RR\text{,}$ because in primary school we were told that $1/0$ is “undefined.” But let's try to define it anyhow.

We note that, as $x$ approaches $0$ from the right, $1/x$ increases without bound; as $x$ approaches $0$ from the left, $1/x$ decreases without bound. If we wanted to add a point that would play the role of $1/0\text{,}$ then this leads us to the following idea: consider the topological space $\RR\sqcup\{\infty\}$ constructed in [cross-reference to target(s) "exm_onepointcompactificationofRR" missing or not unique]. Now we may extend our map $f\colon \RR-\{0\} \to \RR-\{0\}$ to a map $F \colon \RR\sqcup\{\infty\} \to \RR\sqcup\{\infty\}$ by

\begin{equation*} F(x) \coloneq \begin{cases}1/x \amp \text{ if } x \in \RR \smallsetminus \{0\} ; \\ \infty \amp \text{ if } x = 0 ; \\ 0 \amp \text{ if } x = \infty . \end{cases} \end{equation*}

With the topology we've given $\RR\sqcup\{\infty\}\text{,}$ this is continuous! The only thing left for us to check is continuity at $0$ and $\infty\text{.}$

To do this, let $S \subseteq \RR \sqcup \{ \infty \}$ be a subset such that $0$ is close to $S\text{.}$ Let $N >0\text{.}$ There exists an element $s \in S$ such that $|s| \lt 1/N\text{.}$ Thus $|F(s)|>N\text{.}$ It follows that $F$ is continuous at $0\text{.}$

The proof of contiuity at $\infty$ is similar.

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Proposition 2.2.10.

Let $X$ and $Y$ be topological spaces, and let $x \in X .$ The following are equivalent for a map $f : X \to Y .$

The map $f$ is continuous at $x .$
For any open neighborhood $V$ of $f (x),$ the inverse image $f^{- 1} (V)$ is an open neighborhood of $x .$

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Proof.

Assume (1); we aim to prove (2). Let $V$ be an open neighborhood of $f(x)\text{.}$ Let $S \coloneq f^{-1}(Y \smallsetminus V)\text{.}$ If $x \in \tau_X(S)\text{,}$ then

\begin{equation*} f(x) \in \tau_Y(f(S)) \subseteq \tau(Y \smallsetminus V) = Y \smallsetminus V\text{,} \end{equation*}

since $Y \smallsetminus V$ is closed. This is a contradiction, which shows that $x \notin \tau_X(S)\text{.}$ Now let $U \coloneq X \smallsetminus \tau_X(S)\text{.}$ This is now an open neighborhood of $x\text{,}$ and $f(U) \subseteq V\text{.}$

Conversely, assume (2); we aim to prove (1). Let $S \subseteq X$ be a subset, and let $x \in \tau_X(S)\text{.}$ Let $V \coloneq Y \smallsetminus \tau_Y(f(S))\text{.}$ If $f(x) \in V\text{,}$ then by assumption there exists an open neighborhood $U$ of $x$ such that $f(U) \subseteq V\text{.}$ Since $S$ is disjoint from $f^{-1}(V)\text{,}$ it follows that the closure $\tau_X(S)$ is disjoint from $U$ as well. This is a contradiction that implies that $f(x) \notin V\text{,}$ so that $f(x) \in \tau(f(S))\text{.}$

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The definition of homeomorphism is also just the same as for subspaces of Euclidean space:

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Definition 2.2.11.

Let $X$ and $Y$ be topological spaces. A homeomorphism $f : X \to Y$ is a continuous bijection whose inverse $f^{- 1}$ is continuous. The two topological spaces $X$ and $Y$ are homeomorphic if and only if there exists a homeomorphism $X \to Y .$ In this case, we may write $X ≅ Y .$

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Example 2.2.12.

Let's show that the following three topological spaces are homeomorphic:

The circle $S^1 \subseteq \RR^2\text{.}$
The topological space $\PP^1_{\RR}$ from Example 2.1.1.
The topological spaces $\RR \sqcup \{\infty\}$ from Example 2.1.11.

The map $L \colon S^1 \to \PP^1_{\RR}$ defined above is not a homeomorphism, but that doesn't mean we can't find another homeomorphism between these two topological spaces. So let's try to go the other way. Let's define a map $h \colon \PP^1_{\RR} \to S^1\text{.}$ For every line $L \in \PP^1_{\RR}\text{,}$ there is a point $z \in S^1$ such that $L=L(z)\text{.}$ If we think of $S^1 \subset \CC\text{,}$ we can define $h(L)$ as $z^2$ for any $z \in S^1$ such that $L = L(z)\text{.}$ In other words, if $L$ is the line spanned by a nonzero vector $(x,y) \in \RR^2\text{,}$ then

\begin{equation*} h(L) = \left( \frac{x^2-y^2}{x^2+y^2}, \frac{2xy}{x^2+y^2} \right)\text{.} \end{equation*}

This is well-defined, since this formula gives the same value if you replace $(x,y)$ with $(\alpha x, \alpha y)$ for $\alpha \in \RR \smallsetminus \{0\}\text{.}$

Let us prove that $h$ is continuous. We may be tempted to use the fact that the formula above defines a continuous function $\RR^2 \smallsetminus \{0\} \to S^1\text{,}$ but we have to be a little careful, because $\PP^1_{\RR}$ wasn't defined as a subspace of a Euclidean space. Assume that $S \subseteq \PP^1_{\RR}\text{,}$ and assume that $L \in \tau(S)\text{.}$ Let $\varepsilon >0\text{;}$ there exists a line $L' \in S$ such that either $\|i(L) - i(L')\| \lt \varepsilon/2$ or $\|i(L) - j(L')\| \lt \varepsilon/2\text{.}$ Furthermore, we know that no two points in $S^1$ are separated by a distance of more than $2\text{;}$ therefore, both $\|i(L) - i(L')\| \leq 2$ or $\|i(L) - j(L')\| \leq 2\text{.}$ We may therefore multiply these inequalities to obtain:

\begin{equation*} \|i(L) - i(L')\| \|i(L) - j(L')\| \lt \varepsilon\text{.} \end{equation*}

Since $i(L') = -j(L')\text{,}$ we therefore obtain

\begin{equation*} \|h(L) - h(L')\| = \|i(L)^2 -j(L')^2\| \lt \varepsilon\text{.} \end{equation*}

It therefore follows that $h(L)$ is close to $h(S)\text{.}$ So $h$ is continuous.

Now let us show that $h$ is a bijection. If $w \in S^1\text{,}$ then the inverse image $h^{-1}\{w\}$ is the set of lines $L \in \PP^1_{\RR}$ such that either $i(L)$ or $j(L)$ is a square root of $w\text{.}$ In fact, since $i(L) = -j(L)\text{,}$ it follows that $i(L)$ is a square root of $w$ if and only if $j(L)$ is a square root of $w\text{.}$ Hence $h^{-1}\{w\}$ consists of exactly one line: the line passing through either square root of $w$ and the origin.

Now let's demonstrate that $h$ is a homeomorphism. For this, we note that the inverse $h^{-1}$ carries $w \in S^1$ to the line passing through the square roots of $w$ and the origin. To prove that this is continuous, assume that $T \subseteq S^1\text{,}$ and assume that $w \in S^1$ is close to $T\text{.}$ Let $\varepsilon>0\text{;}$ then there exists $w' \in T$ such that $\|w-w'\| \lt \varepsilon^2\text{.}$ Now if $z, -z \in S^1$ are the two square roots of $w\text{,}$ and if $z', -z' \in S^1$ are the two square roots of $w'\text{,}$ then either $|z-z'|\lt \varepsilon$ or $|z+z'|\lt \varepsilon\text{,}$ since otherwise, we would have

\begin{equation*} |w-w'| = |z-z'| |z+z'| \geq \varepsilon^2\text{.} \end{equation*}

Consequently, it follows that $h^{-1}(w)$ is close to $h^{-1}(T)\text{.}$ This now completes the proof that $\PP^1_{\RR}$ and $S^1$ are homeomorphic.

Now we prove that $\RR \sqcup \{\infty\}$ and $S^1$ are homeomorphic. Indeed, define maps

\begin{equation*} f \colon S^1 \to \RR \sqcup \{\infty\} \andeq g \colon \RR \sqcup \{\infty\} \to S^1 \end{equation*}

by the formulas

\begin{equation*} f(x,y) \coloneq \begin{cases}x/(1-y) \amp \text{ if } y \neq 1 ; \\ \infty \amp \text{ if } y = 1 . \end{cases} \end{equation*}

and

\begin{equation*} g(t) \coloneq \begin{cases}1/(t^2+1)(2t, t^2-1) \amp \text{ if } t \neq \infty ; \\ (0,1) \amp \text{ if } t = \infty . \end{cases} \end{equation*}

A direct check confirms that $f$ and $g$ are inverses, but the interesting thing to prove is that they are each continuous.

For $f\text{,}$ continuity away from the point $(0,1)$ follows from the elementary analytic facts we are happy to assume here. But continuity at $(0,1)$ is more interesting! So assume that $S \subseteq S^1$ is a subset to which $(0,1)$ is close. We have to show that $\infty$ is close to $f(S)\text{.}$ So let $N > 0$ be a real number; we aim to find an element $(x,y) \in S$ such that $|x/(1-y)| > N$ or, equivalently, that $x^2/(1-y)^2 = (1+y)/(1-y) > N^2\text{,}$ Choose $\varepsilon>0$ so that $\varepsilon \leq 2/N^2\text{;}$ since $(0,1)$ is close to $S\text{,}$ there exists a point $(x,y) \in S$ such that $x^2 - (1-y)^2 \lt \varepsilon$ and $y \geq 0\text{.}$ Since $x^2+y^2 = 1\text{,}$ this implies that $1-y \lt \varepsilon/2\text{,}$ and since $1+y \geq 1\text{,}$ we obtain

\begin{equation*} \frac{1+y}{1-y} \geq \frac{1}{1-y} > \frac{2}{\varepsilon} > N^2\text{.} \end{equation*}

Once again, elementary analysis facts imply that $g$ is continuous at every point of $\RR\text{.}$ We must show that $g$ is continuous at $\infty$ as well. For this, assume that $T \subseteq \RR \sqcup \{\infty\}$ is a subset such that $\infty$ is close to $T\text{.}$ Let $\varepsilon >0\text{;}$ we aim to show that there exists an element $t\in T$ such that $\|g(t)-(0,1)\| \lt \varepsilon\text{.}$ If $\infty \in T\text{,}$ then this is immediate, so it suffices to consider the case in which $T \subseteq \RR$ is an unbounded subset. There exists $t \in T$ such that $|t| > 2/\varepsilon\text{,}$ and so

\begin{equation*} \left\|1/(t^2+1)(2t, t^2-1) - (0,1)\right\|^2 = \frac{4}{t^2+1} \lt \varepsilon^2\text{,} \end{equation*}

so that $\| g(t) - (0,1) \| \lt \varepsilon\text{,}$ just as we'd hoped.

The upshot here is that we have

\begin{equation*} \PP^1_{\RR} \cong S^1 \cong \RR \sqcup \{\infty\}\text{.} \end{equation*}