I know a map is continuous if the inverse image of an open set is itself open so can you suggest a map between topological spaces that is not continuous?
TIA,
Chad
Do you mean an example of a function between topological spaces which isn't continuous?
If so, then any map $\displaystyle f: \mathbb{R} \to \mathbb{R}$ which isn't continuous under the topology induced by the euclidean metric is what you're asking for.
I mean to say that as a topological space is formed by a collection of subsets satisfying certain rules...
the carrier set is 'the collection of objects from which the the objects making up the subsets are drawn'.
How would you describe those objects?
The collection of subsets you're talking about is the topology of the space.
What you're referring to as the 'carrier set' is the space itself.
Let $\displaystyle X=\{ a , b \}$ and $\displaystyle \tau = \{ X , \phi , \{ a \} \}$ so that $\displaystyle (X, \tau)$ is a topological space.
Let $\displaystyle f:X \to X$ be defined by $\displaystyle f(a) = b , \ f(b)=a$.
Then $\displaystyle f^{-1}( \{a \} ) = \{ b \}$ thus f is not continuous.
Ok, so I can see intuitively why the function...
f(x) = { x+1 if x >
.........{ x if x < 0
is not continuous since you have to raise your pencil in order to draw the graph. But when it comes to using the more formal definition- the only open set we have in the codomain is the set of (strictly) negative values for f(x) so for this function to be not continuous the inverse image of these values (being the only open set) should not be open but... err... it is! So, any clues?
I have read a set can be open and closed at the same time- is that it?
Thanks,
Chad.
I really don't understand what you're trying to say there...
If you're talking about $\displaystyle f:\mathbb{R} \to (-\infty,0) \cup [1,\infty),$
$\displaystyle f(x) = \left\{
\begin{array}{lr}
x+1 & x \geq 0 \\
x & x < 0
\end{array}
\right$
Then simply note that $\displaystyle [1,2)$ is an open subset of $\displaystyle (-\infty,0) \cup [1,\infty )$ (with the usual topology) and that $\displaystyle f^{-1}( [1,2) ) = [0,1)$ which is not open in $\displaystyle \mathbb{R}$ under the usual topology.
I see- that is what I'm driving at.
Now, if I had been asked what is the definition of an open set I would have said it is a set that does not contain its own boundary. I can certainly see that [1,2) is open in the sense of it being a subset that is a topology of the topological space despite being only half open (and half closed) in \mathbb{R}
...but to define an open set as being a thing that is a topology of the topological space would create a circular definition since a topological space is made up of open sets.
So how would you define an open set?
I was thinking about this. I presume it is the case that we can include 1 in the subset even though it is the lower boundary of the subset because the values on the lower side of the boundary that are arbitrary close are not part of the toplogical space.
...but how do you say all this in posh language?
Defunkt, I think you are a very generous person to offer so much help to a complete stranger over the web!
I would disagree with your terminology. The "space" is the set and the topology. I would refer to the underlying set itself as the "base set" but I would have no problem with "carrier set". (Or with "underlying" set!)
Chad950, by if by "discrete and finite" you mean a finite set with the discrete topology then, in the discrete topology, every set is open so if f:A--> B with A discrete, the inverse image of any set is a subset of A, therefore open. Every function in that case is continuous.
As for "open and closed" sets (also called "clopen"), One can show that a set is both open and closed subset of a topological space if and only if it is a "component" in the sense of connectedness (A nonempty set, A, is both open and closed if and only if there exist no larger connected set that contains it as a proper subset).
For example, the only sets that are both open and closed in the Euclidean spaces, $\displaystyle R^n$, are the empty set and the entire set. But if we have, say $\displaystyle A= \left(-\infty, -1\right]\cup \left[1, \infty\right)$, then the empty set, A itself (the empty set and the entire base set are always both open and closed), $\displaystyle \left(-\infty, -1\right]$, and $\displaystyle \left[1, \infty\right)$.
The definition of a topological space is a coupling $\displaystyle (X, \tau)$, where X is your space and $\displaystyle \tau$ is the set of open sets in X, which holds 3 conditions. $\displaystyle \tau$ is called the topology of X. The same set X can be regarded as different topological spaces if you equip it with different topologies.
The definition you are talking about is the definition of an open set in a metric space! In there, an open set is defined as a set $\displaystyle F \subseteq X$ in which, for every $\displaystyle x \in F$ there exists an $\displaystyle \epsilon>0$ such that $\displaystyle B(x, \epsilon) \subseteq F$.
Since there is generally no notion of distance in a topological space, we define open sets by the sets that are members of $\displaystyle \tau$.
In this specific case, we consider $\displaystyle X = (-\infty,0) \cup [1,\infty)$ with the topology induced by the euclidean metric. This means that any open set is a union of open balls under the euclidean metric.I was thinking about this. I presume it is the case that we can include 1 in the subset even though it is the lower boundary of the subset because the values on the lower side of the boundary that are arbitrary close are not part of the toplogical space.
...but how do you say all this in posh language?
Now, it's obvious that for any $\displaystyle x \in (1, \infty)$ there's an open ball containing x which is a subset of $\displaystyle [1, \infty)$. So our problem is only 1. Can you think of an open ball around 1 which is a subset of $\displaystyle [1, \infty)$ in X?
You're welcomeDefunkt, I think you are a very generous person to offer so much help to a complete stranger over the web!
That's the terminology my instructor used . Perhaps I'm getting confused with the translation. Thanks for the correction, though.Originally Posted by HallsofIvy