Continuity

[Xerxes]

I am thoroughly ashamed to be asking such an elementary question, but WTF. I'll start at the beginning, for want of a better place to start......

 

Suppose that [math]S[/math] is a point set, and that [math]\mathcal{P}(S)[/math] is the powerset on [math]S[/math]. Then one defines a topology [math]\tau[/math] on [math]S[/math] by [math]\tau \subseteq \mathcal{P}(S)[/math], and ones calls the pair [math]S(\tau)[/math] as a topological space (though usually the parenthetical \tau is omitted in favour of the assertion that we are dealing with a top. space).

 

Elements in [math]\tau[/math] are called the open sets in [math]S(\tau)[/math], and elements in the complement [math]\tau^c[/math] are closed sets (of course it may happen that a set is both open and closed).

 

Right. Let's drop the \tau, and assert that [math]S,\,\,T[/math] are top. spaces. Then the mapping [math]f:S \to T[/math] is said to be continuous iff, for every set [math]U \in S[/math] that whenever [math]f(U)[/math] is open in [math]T[/math] then [math]f^{-1}(U)[/math] is open in [math]S[/math] (this is the preimage BTW, not necessarily the inverse mapping).

 

So we have that open sets map to open sets under continuous mappings, roughly speaking. So my first question, to which I am pretty sure I know the answer: It seems to me that this concept can be restated in terms of closed sets, id est continuous mappings send closed sets to closed sets. Correct or not? (leaving aside pathological topologies like the discrete or the trivial, for example).

 

Now suppose that the set [math]S =V[/math] is a vector space, and insist that this will become a topological vector space iff

 

a) singletons in [math]V(\tau)[/math] are closed, id est [math]\{x\} \notin \tau[/math]

 

b) vector addition and scaling are continuous in the above sense.

 

Umm..... assuming the standard topology on the vector space [math]\mathbb{R}^1[/math] (again dropping the parentheical \tau) id est the union of all open sets of the form [math](a,b)[/math], and write, for example, vector addition as [math]+:\mathbb{R}^1 \times \mathbb{R}^1 \to \mathbb{R}[/math] so that for any [math]x,\,\,y \in \mathbb{R}^1[/math] that [math](x,y) \in \mathbb{R}^1 \times \mathbb{R}^1 \mapsto x+y = \{z\} \in \mathbb{R}^1[/math].

 

Now if singletons are closed in [math]\mathbb{R}^1[/math] then presumably the ordered pairs [math](x,y)[/math] are likewise closed in the Cartesian product - or is this false reasoning? If not, then addition on the real numbers (as a vector space) is continuous since it sends closed sets to closed sets?

 

Which quite unnecessarily long post brings me back to my point. First is it true that ordered pairs of closed sets in the Cartesian product of vector spaces are closed in the product topology, and second is it reasonable to define continuity in terms of closed sets rather than the more usual in terms of open sets? Third, in this particular example, the "mapping" called addition is clearly surjective (there are more ways to the woods than one!) but is is true in general?

 

Or am I way out to lunch?

 

PS by edit: On proofing this I see I can answer my own questions. Yes, yes, yes and quite possibly. But I shall let it stand, as it may interest someone somewhere out there

Edited June 12, 2011 by Xerxes

[ajb]

The entry on continuous maps in nLab may help you. In particular there is a notion of proper, open and closed continuous maps. I direct you to the nLab.

[DrRocket]

So we have that open sets map to open sets under continuous mappings, roughly speaking.

 

ajb's advice is good.

 

However, it might help you if itn is pointed out that your statement above is wrong. Such maps are called "open maps" and not all continuous maps are open.

 

Continuous -- the inverse image of open sets is open

 

Open -- the direct image of open sets is open

 

Exercise -- find an example of an open map that is not continuous.

 

Recommendation: Get a good course in point set topology under your belt before you tackle functional analysis and topological vector spaces.

Edited June 12, 2011 by DrRocket

[Xerxes]

I thank you both for your helpful comments, but note I used the qualifier "roughly speaking" when I said that continuous mappings send open sets to open sets.

 

I confess I am too lazy to find the example requested by DrRocket, but will instead offer this:

 

Suppose that [math]S,\,\,T[/math] are topological spaces and that [math]f:S \to T[/math]. Let [math]U \in S[/math] be open. Then [math] f[/math] is an open mapping if [math]f(U) =V \in T[/math] is open.

 

Suppose now that [math]f[/math] is also continuous, that is the preimage [math]f^{-1}(V) \in S[/math] is open. This implies there may be some continuous open mapping, say [math]g:T \to S[/math] such that [math]g \cdot f = id_S[/math] and [math]f \cdot g = id_T[/math]. In this circumstance one says that these 2 functions are mutual continuous inverses, and that our spaces are homeomorphic, [math]S \simeq T[/math].

 

Now homeomorphic topological spaces are topologically equivalent, which (again speaking loosely) means that whatever I "do" to one must carry over to the other under the homeomorphic mapping.

 

So I now consider the top. space [math]\mathbb{R}^1[/math] with the standard topology, and assert that whenever the set [math]U \in \mathbb{R}^1 [/math] is open then so is the set [math]U \times U \in \mathbb{R}^1 \times \mathbb{R}^1 \equiv \mathbb{R}^2[/math].

 

I next consider the spaces [math]\mathbb{R}^1\setminus \{0\}[/math] and [math]\mathbb{R}^1\setminus \{0\} \times \mathbb{R}^1 \setminus \{0\} \equiv \mathbb{R}^2 \setminus \{0,0\}[/math].

 

Now [math]\mathbb{R}^1\setminus \{0\}[/math] is not connected, whereas [math]\mathbb{R}^2 \setminus\{0,0\}[/math] is connected, and conclude that there can be no homeomorphism [math]\mathbb{R}^1 \simeq \mathbb{R}^2[/math], since connectedness is a topological property that is preserved under topological equivalence

 

The upshot being that I may have an open mapping between topological spaces which is not a homeomorphism, which implies that not all open mappings are continuous.

[DrRocket]

I thank you both for your helpful comments, but note I used the qualifier "roughly speaking" when I said that continuous mappings send open sets to open sets.

 

I confess I am too lazy to find the example requested by DrRocket,

 

An easy one would be [math] id: \mathbb R \rightarrow \mathbb R[/math] with the usual topology in the domain and the discrete topology in the range.

 

 

 

 

but will instead offer this:

 

Suppose that [math]S,\,\,T[/math] are topological spaces and that [math]f:S \to T[/math]. Let [math]U \in S[/math] be open. Then [math] f[/math] is an open mapping if [math]f(U) =V \in T[/math] is open.

 

Suppose now that [math]f[/math] is also continuous, that is the preimage [math]f^{-1}(V) \in S[/math] is open. This implies there may be some continuous open mapping, say [math]g:T \to S[/math] such that [math]g \cdot f = id_S[/math] and [math]f \cdot g = id_T[/math]. In this circumstance one says that these 2 functions are mutual continuous inverses, and that our spaces are homeomorphic, [math]S \simeq T[/math].

 

Now homeomorphic topological spaces are topologically equivalent, which (again speaking loosely) means that whatever I "do" to one must carry over to the other under the homeomorphic mapping.

 

So I now consider the top. space [math]\mathbb{R}^1[/math] with the standard topology, and assert that whenever the set [math]U \in \mathbb{R}^1 [/math] is open then so is the set [math]U \times U \in \mathbb{R}^1 \times \mathbb{R}^1 \equiv \mathbb{R}^2[/math].

 

OK so far, though the language is a bit stilted. You have the concept of a homeomorphism understood.

 

I next consider the spaces [math]\mathbb{R}^1\setminus \{0\}[/math] and [math]\mathbb{R}^1\setminus \{0\} \times \mathbb{R}^1 \setminus \{0\} \equiv \mathbb{R}^2 \setminus \{0,0\}[/math].

 

 

[math]\mathbb{R}^1\setminus \{0\} \times \mathbb{R}^1 \setminus \{0\}[/math] is not homeomorphic to [math]\mathbb{R}^2 \setminus \{0,0\}[/math].

 

It is homeomorphic to [math] \mathbb R \times \mathbb R \setminus \{(\mathbb R \times \{0 \}) \cup (\{0 \} \times \mathbb R)\}[/math], which is not connected.

 

 

 

Now [math]\mathbb{R}^1\setminus \{0\}[/math] is not connected, whereas [math]\mathbb{R}^2 \setminus\{0,0\}[/math] is connected, and conclude that there can be no homeomorphism [math]\mathbb{R}^1 \simeq \mathbb{R}^2[/math], since connectedness is a topological property that is preserved under topological equivalence

 

It is true that [math]\mathbb{R}^1[/math] is not homeomorphic to [math] \mathbb{R}^2[/math] but that does not follow from your argument.

 

It is also true that [math]\mathbb{R}^2 \setminus\{0,0\}[/math] is connected and therefore cannot be mapped continuously onto [math]\mathbb{R}^1\setminus \{0\}[/math] which is not connected. What might surprise you is that [math]\mathbb{R}^1\setminus \{0\}[/math] can be mapped continuously onto [math]\mathbb{R}^2 \setminus\{0,0\}[/math] (this is far from obvious and requires theorems on "space filling curves").

 

 

The upshot being that I may have an open mapping between topological spaces which is not a homeomorphism, which implies that not all open mappings are continuous.

 

This does not seem to follow from even the fact that no two of [math]\mathbb{R}^1[/math], [math] \mathbb{R}^2[/math], [math]\mathbb{R}^1\setminus \{0\}[/math] and [math]\mathbb{R}^2 \setminus\{0,0\}[/math] are homeomorphic.

 

Moreover it is quite easy to have an open mapping that is not a homeomorphism, say [math] (x,y) \rightarrow x [/math] from [math] \mathbb R^2[/math] to [math] \mathbb R [/math]. How this implies that open maps need not be continuous (a true statement) is not evident.

 

 

As an aside you may be interested in theinvariance of domain theorem for maps between Euclidean spaces of like dimension that shows that injective continuous maps are open.