Understanding the definition of continuous functions

[Robin04]

Definition: A function f mapping from the topological space X to the topological space Y is continuous if the inverse image of every open set in Y is an open set in X.

The book I'm reading (Charles Nash: Topology and Geometry for Physicists) emphasizes that inversing this definition would not be a sufficient requirement for continuity: a function that maps open sets into open sets is not necesserily continuous (according to our intuition). He presents a counter example $f(x) = \begin{cases} x& x\leq 0 \\ 1+x & x > 0 \end{cases}$

He says that this is a function that maps open sets into open sets, but the inverse image of an open set is not necesserily an open set. The second part is trivial if we look at the image of the interval $(1- \epsilon, 1+\epsilon)$, so according to the real definition this isn't a continuous function. But I'm having a problem with the first part. If we take the interval $(-\epsilon,\epsilon)$, then its image is $(-\epsilon, 0] \cup(1,1+\epsilon)$ This is not an open interval either. What am I missing here?

 

[fresh_42]

Definition: A function f mapping from the topological space X to the topological space Y is continuous if the inverse image of every open set in Y is an open set in X.

The book I'm reading (Charles Nash: Topology and Geometry for Physicists) emphasizes that inversing this definition would not be a sufficient requirement for continuity: a function that maps open sets into open sets is not necesserily continuous (according to our intuition). He presents a counter example $f(x) = \begin{cases} x& x\leq 0 \\ 1+x & x > 0 \end{cases}$

He says that this is a function that maps open sets into open sets, but the inverse image of an open set is not necesserily an open set. The second part is trivial if we look at the image of the interval $(1- \epsilon, 1+\epsilon)$, so according to the real definition this isn't a continuous function. But I'm having a problem with the first part. If we take the interval $(-\epsilon,\epsilon)$, then its image is $(-\epsilon, 0] \cup(1,1+\epsilon)$ This is not an open interval either. What am I missing here?

It is an open set in $Y:=(-\infty,0] \cup (1,\infty)$. If you consider $f\, : \,\mathbb{R} \longrightarrow \mathbb{R}$ instead, then you fill in the unnecessary gap $(0,1]$. However, we can define $f\, : \, \mathbb{R}\longrightarrow (-\infty,0] \cup (1,\infty)$ in which case $f$ is an open function. And it is still not continuous at $x=0$.

 

[Robin04]

It is an open set in $Y:=(-\infty,0] \cup (1,\infty)$. If you consider $f\, : \,\mathbb{R} \longrightarrow \mathbb{R}$ instead, then you fill in the unnecessary gap $(0,1]$. However, we can define $f\, : \, \mathbb{R}\longrightarrow (-\infty,0] \cup (1,\infty)$ in which case $f$ is an open function. And it is still not continuous at $x=0$.

So the reason why its continuous is because the range of $f$ is only a subset of $\mathbb{R}$ and thus the interval $(-\epsilon,0] \cup (1,1+\epsilon)$ is open because I cannot take $0+\delta$, $\delta > 0$ as it's not in the set? It would only be closed if $0+\delta$ was in the set?

 

[fresh_42]

Not sure what you mean here. Let's keep the topological definition. Since it makes a statement about open sets of $X$ and $Y$, the choice of the spaces is essential here, and $Y=(-\infty,0]\cup (1,\infty) \neq \mathbb{R}$. Next we see, that if we concentrate on the open sets of $Y$ instead of the different space $\mathbb{R}$, then $f$ becomes open. Now $U:=(-\varepsilon,0]\subseteq Y$ is an open set and $f^{-1}(U)=(-\varepsilon,0] \subseteq X:= \mathbb{R}$ is not, so it cannot be continuous.

Now you want to switch to the equivalent $\varepsilon-\delta$ definition of continuity in $\mathbb{R}$. Let us therefore take $\varepsilon := 1/2$. Now we must find a $\delta > 0$ such that for all $x\in \mathbb{R}$ holds $|x| < \delta \Longrightarrow |f(x)|<1/2\,.$ But such a $\delta$ does not exist, because for $x_0:=\frac{1}{2}\delta$ we sure have $|x_0|=\frac{1}{2}\delta < \delta$ so our condition is given, but $|f(x_0)|=1+\frac{1}{2}\delta > 1/2\,.$

 

[lavinia]

Here is an extreme example.

Let $X$ be the real numbers with the usual topology. Let $Y$ be the real numbers with the discrete topology. Let $f:X→Y$ be the identity map $f(x)=x$ for all numbers. Then $f$ is an open mapping because every subset of $Y$ is open. But the inverse image of an open set might not be open for instance the inverse image of a point.

Now let $X$ have the trivial topology - the only open sets are the null set and the set of all real numbers - and $Y$ the usual topology on the real numbers. Then the identity map is open but not continuous. More generally, $Y$ can be the real numbers with any non-trivial topology.

 

[Robin04]

It's become a bit complicated for me. I'm very early in my studies in topology. What I understood is that $f$ does not map from $\mathbb{R}$ to $\mathbb{R}$, but from $\mathbb{R}$ to $Y:=(-\infty, 0] \cup(1, \infty)$. There was the example I checked where the function mapped an open set from $\mathbb{R}$ into $(-\epsilon, 0] \cup(1,1+\epsilon)$. The only thing to prove is that this set is open in $Y$. First, I think I'm missing the exact definition of what an open set generally is. The book I'm reading only says that open sets are the elements of a topology to a set, but an open set also seems to be a separate concept used outside the context of topological spaces or at least not very much attached to them (correct me if I'm wrong).

Also, you both used the term open for functions too. When do we say that a function is open?

 

[fresh_42]

It's become a bit complicated for me. I'm very early in my studies in topology. What I understood is that $f$ does not map from $\mathbb{R}$ to $\mathbb{R}$, but from $\mathbb{R}$ to $Y:=(-\infty, 0] \cup(1, \infty)$. There was the example I checked where the function mapped an open set from $\mathbb{R}$ into $(-\epsilon, 0] \cup(1,1+\epsilon)$. The only thing to prove is that this set is open in $Y$. First, I think I'm missing the exact definition of what an open set generally is.

An open set is what is defined as being open in the corresponding topology.

A topology is a set $X$ together with a collection of subsets $U_\iota \in \mathcal{P}(X)$ with certain conditions. Now let this collection of subsets be $\tau = \{\,U_\iota\,|\,\iota \in I\,\}$ for some index set $I$. Then $\tau$ is called a topology if

• $\emptyset\; , \; X \in \tau$
• Finite intersections of sets of $\tau$ are in $\tau$ again: $U_\mu\; , \;U_\nu \in \tau \Longrightarrow U_\mu \cap U_\nu \in \tau$
• Arbitrary unions of sets of $\tau$ are in $\tau$ again: $U_\iota \in \tau \,(\iota \in J \subseteq I)\,\Longrightarrow \bigcup_{\iota \in J}U_\iota \in \tau$

The sets in $\tau$ are all sets we call open in the topology $\tau$.

The book I'm reading only says that open sets are the elements of a topology to a set, but an open set also seems to be a separate concept used outside the context of topological spaces or at least not very much attached to them (correct me if I'm wrong).

Also, you both used the term open for functions too. When do we say that a function is open?

You probably confuse different topological spaces here. We start with $X=\mathbb{R}$ and define as open whatever can be written as an arbitrary union of open intervals.

For the sake of completeness: This is the topology which is induced by the Euclidean metric. Every metric induces a topology, because we can define open sets as those with a distance from a point smaller than a fixed number: $U_\iota(x_0;r) := \{\,x \in X\,|\,d(x_0,x)< r\,\}$, collect the empty set, the entire space, and all arbitrary unions of $U_\iota(x_0;r)$ and call them open.

Now we have what you know about the real numbers. Next, we observe that $Y\subsetneq \mathbb{R}$ is a proper subset. Therefore, we cannot take the same sets as in $\mathbb{R}$ as open sets, simply because some of them aren't part of $Y$. However, we can introduce the so called subspace topology, that is we define for $(Y,\tau)$ all sets $U \cap Y \in \tau$ for some open set $U \in (\mathbb{R},|.|)$ where $|.|$ means the topology induced by the usual metric.

Thus $(-\varepsilon,0] = (-\varepsilon,1/2) \cap Y \subseteq Y$ is open, because $(-\varepsilon,1/2) \subseteq \mathbb{R}$ is, and likewise $(1,1+\varepsilon)=(1,1+\varepsilon) \cap Y$ which is open in both, $\mathbb{R}$ and $Y$. Now since $(-\varepsilon,0] \cup (1,1+\varepsilon)$ is a union of open sets, it is again open in $Y\,.$

 

[lavinia]

Also, you both used the term open for functions too. When do we say that a function is open?

A function is called open if it maps open sets into open sets.

 

[WWGD]

It's become a bit complicated for me. I'm very early in my studies in topology. What I understood is that $f$ does not map from $\mathbb{R}$ to $\mathbb{R}$, but from $\mathbb{R}$ to $Y:=(-\infty, 0] \cup(1, \infty)$. There was the example I checked where the function mapped an open set from $\mathbb{R}$ into $(-\epsilon, 0] \cup(1,1+\epsilon)$. The only thing to prove is that this set is open in $Y$. First, I think I'm missing the exact definition of what an open set generally is. The book I'm reading only says that open sets are the elements of a topology to a set, but an open set also seems to be a separate concept used outside the context of topological spaces or at least not very much attached to them (correct me if I'm wrong).

Also, you both used the term open for functions too. When do we say that a function is open?

A set S is open if for every x in S there exists a basis element $B_x$ withx$x\in B_x \subset S$. Meaning each point has a basic neighborhood contained entirely in the set.

 

[WWGD]

It's become a bit complicated for me. I'm very early in my studies in topology. What I understood is that $f$ does not map from $\mathbb{R}$ to $\mathbb{R}$, but from $\mathbb{R}$ to $Y:=(-\infty, 0] \cup(1, \infty)$. There was the example I checked where the function mapped an open set from $\mathbb{R}$ into $(-\epsilon, 0] \cup(1,1+\epsilon)$. The only thing to prove is that this set is open in $Y$. First, I think I'm missing the exact definition of what an open set generally is. The book I'm reading only says that open sets are the elements of a topology to a set, but an open set also seems to be a separate concept used outside the context of topological spaces or at least not very much attached to them (correct me if I'm wrong).

Also, you both used the term open for functions too. When do we say that a function is open?

When considering Y it is seen as a subpace ( of t Reals, here) so that open (subspace)sets are intersections of open sets in the ambient space ( the Reals in this case) with the subspace ( here described by freshmeister in post #2.).