Continuous Map if Both Are Given the Corresponding Inductive Limit Topology Stackexchange

October 12, 2022 Post a Comment

Following what we did for real analysis, we have the following definition of limits.

Definition of Limits. Let X, Y be topological spaces and $a\in X$ . If f : X-{a} → Y is a function, then we write $\lim_{x\to a} f(x) = b \in Y$ if the function:

$g:X\to Y,\ g(x) = \begin{cases} f(x), &\text{if } x\ne a, \\ b, &\text{if }x=a.\end{cases}$

is continuous at x=a. In words, we say that f(x) approaches b as x approaches a.

Definition of Convergence. If X is a topological space, a sequence of elements $x_n \in X$ is said to converge to a if the function f :N* – {∞} → X, f(n) = x_n, approaches a as n approaches ∞. A sequence which has a limit is called aconvergent sequence.

[ Refer here for the definition ofN*. ]

sequence_in_topo

We shall show that this convergence is consistent with our earlier definition of convergence.

Proposition. Let (X, d) be a metric space. Then a sequence $x_n\in X$ converges to a (in the above definition) if and only if:

for any ε>0, there exists N such that whenever n>N, $d(x_n, a) < \epsilon$ .

Proof.

Suppose $x_n\in\mathbf{R}$ converges to a. Given any ε>0, N(a, ε) is an open ball containinga. By definition, if g:N* → X is defined by g(n)=x_n , forn=1, 2, 3, … andg(∞)=a, theng is continuous at ∞. Thus $U:=g^{-1}(N(a,\epsilon))$ contains an open subset ofN* containing ∞. SoU contains someU_N = {N,N+1,N+2, … } and $n>N\implies n\in U\implies x_n=g(n)\in N(a,\epsilon).$
Conversely, assume the condition in the proposition holds; let's prove thatg is continuous at ∞. Now any open subset $V\subseteq X$ containinga must contain an open ball $N(a,\epsilon) \subseteq V$ for some ε>0. Then there existsN such that whenevern>N, $d(x_n, a)<\epsilon \implies x_n \in N(a,\epsilon).$ Thus $g^{-1}(V)$ contains the setU_N ₊₁ = {N+1,N+2,N+3, …, ∞} which is an open subset containing ∞. ♦

In addition toN*, let's consider the extended real line $\overline{\mathbf{R}} = \mathbf{R}\cup \{-\infty, \infty\}.$ Again we should think of the infinities as dummy symbols. Its topology is defined via the basis:

Then all prior limits can be expressed via the extended real line. For example:

Proposition.

Proof.

We'll prove (LHS) → (RHS) and leave the converse to the reader.

Let's prove the first statement. Suppose LHS holds; we need to show $g:\mathbf{N}^* \to \overline{\mathbf{R}}$ which takes $n\mapsto x_n, \infty\mapsto \infty$ is continuous at ∞. Indeed, any open subset of $\overline{\mathbf{R}}$ containing ∞ must contain some (L, ∞]. By classical definition, there existsN such that whenevern>N, we havex_n >L. Then $g^{-1}((L,\infty])$ contains $U_{N+1}\cup\{\infty\}$ , an open set containing ∞.

For the second statement, again assume LHS holds. Let $g:\overline{\mathbf{R}} \to \mathbf{R}$ be the function which takes realx tof(x) and ∞ toL; our job is to prove continuity ofg at ∞. LetV be an open subset ofR containingL, which contains (L-ε,L+ε) for some ε>0. By classical definition, there existsM such that wheneverx>M,f(x) lies in (L-ε,L+ε) and henceV. Thus, $(M,\infty] \subseteq f^{-1}(V)$ is an open subset of $\overline{\mathbf{R}}$ containing ∞, andg is continuous at ∞. ♦

The point we're trying to say is this.

Summary. The various multi-case definitions of limits and convergence can all be subsumed under a single definition by considering various topological spaces.

Basic Properties of Limits

The following is basic.

Proposition. Let f : X-{a} → Y be a function where $\lim_{x\to a} f(x) = b$ . If g : Y → Z is continuous at b, then gf : X-{a} → Z satisfies $\lim_{x\to a} gf(x) = g(b)$ .

Proof.

Define $h:X\to Y$ and $i:X\to Z$ via:

$h(x)=\begin{cases} f(x), &\text{if } x\ne a,\\ b, &\text{if }x=a,\end{cases}$ and $i(x) = \begin{cases} gf(x), &\text{if } x\ne a,\\ g(b), &\text{if }x=a.\end{cases}$

From the condition, we know thath is continuous at a. Since g is continuous atb=h(a),i =gh is also continuous at a. ♦

Corollary. If $x_n \to a$ in the topological space X, and f : X → Y is continuous, then $f(x_n)\to f(a)$ in Y.

Proof

Since $x_n\to a$ , the function $g:\mathbf{N}^* \to X$ which takes $n\mapsto x_n$ has a limit $\lim_{n\to\infty} g(n) = a$ . By the previous proposition, we get $\lim_{n\to\infty} f(g(n)) = f(a)$ and so $f(x_n)\to f(a).$ ♦

Next we shall approach the basic question: is the limit $\lim_{x\to a} f(x)$ unique? Phrased in such generality, the answer is no, as we saw counter-examples even for the case whereX is a subset ofR.

So let's first consider sequences.

Theorem. Let X be a topological space. Assume:

Then every sequence $x_n \in X$ has at most one limit in X. A topological space satisfying the above property is said to beHausdorff.

hausdorff

Thus, the Hausdorff property is a sufficient condition for unique convergence, but it's known that the condition is not necessary.

Proof.

Supposea,b are distinct limits of $(x_n)$ . Pick open subsetsU,V such that $a\in U$ , $b\in V$ , $U\cap V=\emptyset$ . Then there existsM,N such that (i) whenevern>M, we have $x_n \in U$ , (ii) whenevern>N, we have $x_n\in V$ . The two statements clearly contradict. ♦

One obvious source of Hausdorff topological spaces is via metric spaces, i.e. metric spaces are Hausdorff. Indeed, ifx,y are distinct points in a metric spaceX, then letting ε=d(x,y)/2 > 0, we have $N(x,\epsilon) \cap N(y,\epsilon) = \emptyset$ , for ifz satisfiesd(x,z) < ε andd(y,z) < ε, thend(x,y) ≤d(x,z) +d(z,y) < 2ε = d(x,y), which is absurd. In particular, any convergent sequence in a metric space has a unique limit.

For uniqueness of general limits, obviously we have to care about the domain spaceX.

Theorem. Let f : X-{a} → Y be a map of topological spaces. Assume the following.

Y is Hausdorff.

The singleton set {a} is not open in X.

Then there's at most one limit for $\lim_{x\to a} f(x)$ .

Proof.

Suppose $b,c\in Y$ are distinct limits. Hence the functions $g_1, g_2:X\to Y$ ,

$g_1(x) = \begin{cases} f(x), &\text{if } x\ne a,\\ b, &\text{if }x=a\end{cases}$ and $g_2(x)=\begin{cases} f(x), &\text{if }x\ne a,\\ c &\text{if }x=a\end{cases}$

are continuous. Pick open subsetsU andV ofY, such that $b\in U$ , $c\in V$ and $U\cap V=\emptyset$ . Now $g_1^{-1}(U) = f^{-1}(U)\cup\{a\}$ must contain an open subset which containsa. Same for $g_2^{-1}(V) = f^{-1}(V) \cup\{a\}$ . But since $f^{-1}(U)\cap f^{-1}(V) = f^{-1}(U\cap V)=\emptyset$ , the intersection of these two open subsets is {a}, which contradicts the second condition. ♦

Note: it's easy to find a counter-example when {a} is open inX. In an earlier article, we saw the following:

kindacontinuous

Here $X=(-\infty, -1)\cup \{0\}\cup (1, \infty)$ andf :X-{0} →R is defined byf(x)=x. But because {0} is an open subset ofX,f(0) can take any value without violating its continuity.

flinchumstren1944.blogspot.com

Source: https://mathstrek.blog/2013/02/02/topology-limits-and-convergence/

Flinchum Stren1944

Continuous Map if Both Are Given the Corresponding Inductive Limit Topology Stackexchange

Basic Properties of Limits

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