# Tag Info

1

We do it because it works; we have an idea that some things should be continuous and other things not, and with this definition, they are or are not as desired! Ultimately, that's what really matters: that the consequences of the definition match whatever notion we're trying to capture. The definition doesn't have to be intuitive, be natural, or even bear ...

1

Great question! The answer is yes if you restrict your attention to sufficiently nice spaces. Specifically, each of the following statements is true (1 and 3 are quite well-known): Let $Y$ be a Hausdorff space and $X\subseteq Y$ be a compact subspace. Then $X$ is closed in $Y$. Let $X$ be a regular space such that whenever $X$ is a subspace of a ...

-2

The closed unit ball in an infinite dimensional normed space is not compact.

9

No; take the two-point topological space with one open point and one closed point. The open point is compact, but not closed. In fact, I don't think that there are any topological spaces which are closed inside every other topological space.

5

First note that the sequence is bounded: $$|\phi_n(f)|=n^{-1}\,\left|\sum_1^nf(j)\right|\leq\max\{|f(j)|:\ j=1,\ldots,n\}\leq\|f\|_\infty.$$ This shows that $\|\phi_n\|\leq1$ for all $n$, so the sequence $\{\phi_n\}$ lies in the unit ball of $(\ell_\infty)^*$. In the weak$^*$-topology, the unit ball of the dual is compact, and so every sequence within it ...

0

In $\Bbb R^3$, a set is compact if and only if it is closed and bounded, and your set there is not bounded. So it is not compact. Pre-images of compact sets by functions (even continuous ones) need not be compact, that's the mistake in your proof.

5

Consider $f\colon A\to B$. Changes in the domain "control" what happens to the image, the value of $x\in A$ is "causal" for the value $f(x)\in B$. Therefore you may think that going from domain to image is the natural way to define continuity. However, continuity is about how well we can predict the image $f(x)$ if we slightly modify $x$; that is, if only ...

1

The metric space version makes some sense. In the metric space case, given any ball $B_1$ (perhaps very small) around $f (x)$, there is a ball $B_2$ around $y$ where $f (y) \in B_1$. Informally, you have some perhaps small amount of wiggle room in the domain, given any positive maximum amount of wiggle in the codomain. In this sense "small changes in the ...

2

Hint. Use two ingredients to prove the completness of $\mathcal C^2([0,1])$: Ingredient one the space $\mathcal C([0,1]$ of continuous real functions defined on $[0,1]$ is complete for the $\Vert f \Vert =\sup\limits_{x \in [0,1]} \vert f(x) \vert$ norm. Ingredient two if a sequence of differentiable functions $(f_n)$ is such that $f_n^\prime$ converges ...

3

You can formulate continuity in terms of images. A function $f:X\to Y$ is continuous at a point $x$ if and only if for every neighborhood $U$ of $f(x)$ there exists a neighborhood $V$ of $x$ such that $f(V)\subseteq U$.

9

It's not nice to formulate it in terms of images. The behavior of images varies wildly between various continuous functions. Take the interval $(-10,10)$ and the functions $f_1(x) = x$ and $f_2(x) = \sin x$. Then the image of $(-10,10)$ under $f_1$ is of course $(-10,10)$, but the image of $(-10,10)$ under $f_2$ is $[-1,1]$. With one function you get an open ...

4

You are right in not being convinced: the parity of the number of twists is a topological invariant, but the actual number is not. To see this think of the strip as $I \times I$ (where $I = [0, 1]$) with opposite edges $\{0\} \times I$ and $\{1\} \times I$ identified either by $(0, x) \mapsto (1, x)$ (for an even number of twists) or by $(0, x) \mapsto (1, ... 1 I hope that this will make things more clear to you. In this answer$i$corresponds with the$f^*$in your question and$s$with the$\pi_f$in your question. Let$X$and$Y$be topological spaces and let$f:X\rightarrow Y$be continuous. Looking at$f$purely as a function we can write it as$f=i\circ s$where$i$is injective and$s$is surjective. As ... 0 Let$p\in X$be a point and$H\subseteq X$be a closed set such that$p\notin H$. Since$H^c$is open and$p\in H^c$, there is a basic open set$V\subseteq X$such that$p\in V\subseteq H^c$. Case I:$V=(a,\infty)$for some$a\in X$. If there is some$x\in X$such that$a<x<p$, then$(-\infty,x)$and$(x,\infty)$are disjoint open sets containing$H$... 2 First of all I'd like to say that I'm very amused by the fact that you and I independently decided to call wheel theory's$0/0$'nullity' after James Anderson's 'transreal arithmetic'. I'm fairly sure that if you take the real projective line topology on$\mathbb{R}\cup\{\infty\}$and append$\Phi$as an open extension topology (i.e. the open sets are ... 0 No, such a thing cannot exist. If your circloid has nonempty interior, then it will contain an open disk, and then it will in particular contain a circle that fits into this open disk as a proper subset, so it will fail the minimality part of the definition. 11 For example,$\{(x,y)\in\mathbb R^2\mid xy\ge 0\}$(that is, the first and third quadrant of the plane, plus the coordinate axes). This is a star-domain with star center$(0,0)$, but its interior is not even connected, and so cannot be a star-domain. Whenever the original star-domain has an interior point as star-center, its interior will still be a ... 3 Let$Af = \int_{0}^{x}f(t)dt$in$L^{2}[0,1]$. Then$A : L^{2}\rightarrow L^{2}$is bounded. Let$W$consist of all continuously differentiable$g \in L^{2}[0,1]$for which$g(0)=g(1)=0$.$W$is dense in$L^{2}[0,1]$because$\{ \sin(n\pi x) \}_{n=1}^{\infty}\subset W$is an orthogonal basis of$L^{2}[0,1]$. However,$A^{-1}W$is not dense because$f \in ...

0

Use the infinite pigeon-hole principle: If aninfinite set is presented as the union of finitely many subsets,at least one subset is infinite.So if a sequence p(n) converges to p, where each p(n) belongs to some F(j), then,for at least one j, there are infinitely many n for which p(n) belongs to F(j), so p belongs to this F(j).

0

The absolute value of $|a|$ is $a$ or $-a$, depending on whether $a$ is nonegative or not. In your case, the condition $n\geq m$ implies $f_m (t)\geq f_n (t)$, so $$|f_n (t)-f_m (t)|=-(f_n (t)-f_m (t))=f_m(t)-f_n (t).$$ For your second question, when you calculate your integrals you are forgetting the antiderivatives.

2

Uniform closure, of course not. Closure in the topology of pointwise convergence yes, as has been pointed out, but this is trivial (immediate from the fact that a polynomial can take any values you want on any finite set) and I don't see how it's good for anything. There's a closure for which the answer is yes, and which also actually allows one to prove ...

2

Suppose there were such continuous maps. Then $\eta$ would be injective. Hence it's restriction to the closed unit disc is a homeomorphism. The image of that disc would be a closed bounded interval. But the disc is not homeomorphic to an interval: removing an interior point from the interval leaves a disconnected set, while removing any point from the disc ...

1

I would modify the proof a bit. In the first sentence you could write Let $x∈A, ϵ>0$ and $U=(x−ϵ,x+ϵ)$. Since $U∩A$ $\color{red}{\text{is a singleton}}$ or is not connected, $U∩(ℝ∖A)≠∅$ and therefore, $x∉A^∘$. The other direction is fine. I think you don't need the hypothesis that $|A|\ge 2$. If $A$ is a single point, then it's totally disconnected ...

4

A continuous map from $\Bbb R^2$ to $\Bbb R$ cannot be injective. This is trivial; proof below. It's higher-dimensional versions, like the fact that a continuous map from $\Bbb R^3$ to $\Bbb R^2$ cannot be injective, that I suspect are non-trivial. Given two points $a,b\in\Bbb R^2$, let $[a,b]$ denote the line segment from $a$ to $b$. Say $f:\Bbb ... 0 All polynomials are dense in borel functions in pointwise convergence topology because polynomials are dense in continous functions in uniform convergence topology, and those are dense in borel in pointwise convergence topology. However if you restrict to polynomials vanishing at zero you will only get borel functions vanishing at zero. -1 Perhaps, we can suppose that$X$and$Y$are metric spaces (in particular subset of real line$\mathbb R$) and that$f(x,y)$is real-valued function in$x ,y$. For a simple model we can take$X=\mathbb R$and$Y=[a,b]$closed interval in$\mathbb R$. Set$I(x) = \inf_{y\in Y} f(x,y)$. Fix a point$x_0$in$X$. Let us prove that ... 2 No, it’s not true that every closed set in$\Bbb R$with the lower limit topology is a union of sets of the form$\Bbb R\setminus[a,b)$. For example,$\{0,1\}$is a closed set, because its complement is the open set $$\bigcup_{x<0}[x,0)\cup\bigcup_{0<x<1}[x,1)\cup\bigcup_{x>1}[x,x+1)\;.$$ Let$\tau$be the lower limit topology. It’s not hard to ... 0 Open sets are going to look like arbitrary unions of$[a,b)$. Closed sets are defined to be the complement of an open set. So if$U$is open, $$U = \bigcup_{i \in I} {[a_{i}, b_{i})}.$$ Then, $$X - U = X - \bigcup_{i \in I} {[a_{i},b_{i})} = \bigcap_{i \in I} {[a_{i},b_{i})^{C}}$$ So your closed sets are looking like intersections of the sets you mentioned ... 0 Sleeping on it helps. :-) A sufficient conditions for wild functions to be dense is that there exists one wild function$X\to\mathbb N\times Y$. Calling this function$g$, we can approximate any$f$by $$f_n(x) = \begin{cases} y &\text{when }g(x)=\langle k,y\rangle\text{ with }k>n \\ f(x) &\text{otherwise} \end{cases}$$ In the$\mathbb ...

1

The integral (also known as tha $L^1$ metric) and the uniform metric cannot be defined on any space. The uniform metric is defined on sets of bounded, real (or complex) valued functions defined on a set $X$. If $f,g\colon X\to\mathbb{R}$ are bounded, then $$d_\text{u}(f,g)=\max_{x\in X}|f(x)-g(x)|.$$ It measures the greates difference between the values of ...

2

Here's a hint. The continuous image of a path connected set is path connected, so you can try to find a relevant map.

6

If $(x_1,\sin(1/x_1))$ and $(x_2,\sin(1/x_2))$ are two points of $A$, then they can be connected by the path $$\gamma(t)=\left(t,\sin\left(\frac1t\right)\right),\;t\in[x_1,x_2]$$

1

I know from your other questions you like to see rigorous answers; so I thought I'd do the same on this one too. I did this a slightly longer way since I do not know which definition of closed you are using (so I used a standard definition of open) def. $U$ is open $\Leftrightarrow \forall x \in U\; \exists \varepsilon >0 : B(x, \varepsilon) \subset U$ ...

0

Given such an $A$, let $a\sim b$ if $a=b$ or $a$ is the successor of $b$ or $b$ is the successor of $a$. Your hypotheses imply this is an equivalence relation; let $B=A/{\sim}$ be the quotient. Then $B$ is a countable bounded dense linear order. Let $C\subseteq B$ be the set of equivalence classes that contain two points rather than one. Then your ...

2

Corrected. Not quite. First, it can be isomorphic to $\Bbb Q\cap(\mathscr{C}\cup F)$ for any $F\subseteq\{-1,2\}$, since the first point can have a successor, and the last can be a successor. More important, the third condition isn’t quite strong enough to do what you want. As Eric Wofsey reminded me in the comments, your conditions allow $A$ to be $(\Bbb ... 0 For example if X= {0, 1, 2} with d(x, y) defined as d(x, y)= |x- y| then there are only 2^3= 8 subsets so at most 8 "open balls". 0 It is not clear to me what you are asking exactly. Given a ball$B=B(x;r)$in$\Bbb R^n$all the balls$B(x;\rho)$with$\rho<r$are inside$B$and so in whatever open set$A$the ball$B$is a subset of. This is entirely trivial. A bit less trivial question would be if you can find infinitely many disjoint balls inside an open set$A\subseteq\Bbb R^n$, ... 2 It may be the case that these open balls are actually equal to each-other despite having different radii. For instance, if we equip$\mathbb{Z}$with the usual metric, then$B_{1/2}(0) = B_{1/3}(0)$, for instance. 8 In any metric space, there are an infinite number of ways to write down balls with a given center. But some of the balls might actually be the same. For instance, in the "discrete metric"$d(x,y)=0$if$x=y$and$1$otherwise, all balls$B_r(x)$for$r \leq 1$are the same (they are just$\{ x \}$) while all balls$B_r(x)$for$r>1$are also the same ... 0 In Functional Analysis, one often speaks of projective limits (limits in the category LCS of locally convex spaces) and inductive limits (colimits in LCS). In the latter case one has to be careful as colimits in TOP usually differ from those in LCS. What you state as a theorem says that every locally convex space is a projective limit of seminormed spaces. ... 2 The ring in question is the ring of rational functions on the projective line over$\mathbb{F}_p$which possibly have poles at$0$and$\infty$. The embedding of$\mathbb{F}_p[T, 1/T]$in question (the projection of the adelic embedding) is taking corresponding Laurent series expansions at$0$and$\infty$. To see that this embedding is discrete, one has ... 4 The argument given in this answer on MO shows that if$A$is any set equipped with a collection of finitary operations which satisfy some equational axioms such that$2\leq|A|\leq\aleph_0$, then there is an infinite-dimensional contractible CW-complex$X$which can be equipped with corresponding finitary operations which are continuous and which satisfy the ... 4 Given a space$E$equipped with a cell decomposition for which it has the weak topology and the boundary of every cell is contained in the union of the lower-dimensional cells, closure-finiteness is equivalent to the statement that for each$n$, the$n$-skeleton$E^n\subseteq E$(i.e., the union of the cells of dimension$\leq n$) has the weak topology. ... 0 You might check the proof of Cartesian coordinates for vertices of a regular 16-simplex? for a formula for the regular simplex vertices 4 The product topology is discrete if and only if$I$is finite. If$I$is countably infinite, the product is homeomorphic to the Cantor set. The complement of a singleton is not a singleton unless$I$is a one-point set. If$I$has at least two elements, then$\prod_{i\in I}\{0,1\}$has at least$2^2=4$elements, so the complement of a singleton has at least ... 0 Your proof is correct, but you’ve not expressed it very well. Here’s a clearer version: If$X$is not connected, there are disjoint non-empty open sets$A$and$B$such that$X=A\cup B$. Since$A$and$B$are non-empty,$X\setminus A$and$X\setminus B$are finite. But$X\setminus A=B$, and$X\setminus B=A$, so$A$and$B$are both finite, and hence so ... 7 HINT: Consider the sequence$\langle x_n:n\in\Bbb N\rangledefined by $$x_n=\begin{cases} 0,&\text{if }n\text{ is even}\\ -\frac1n,&\text{if }n\text{ is odd}\;. \end{cases}$$ 3 For the first highlighted step, note that \begin{align*} \| x - a \| &\le \max(\|x - a\|, \|y - b\|) = \|(x,y) - (a,b)\| \\ \| y - b \| &\le \max(\|x - a\|, \|y - b\|) = \|(x,y) - (a,b)\| \\ \| a \| &\le \max(\|a\|, \|b\|) = \|(a,b)\| \\ \| b \| &\le \max(\|a\|, \|b\|) = \|(a,b)\|. \end{align*} Therefore, \begin{align*} \| B \| \| x - a \| \| ... 2 Maybe you can try using sequential continuity, equivalent to continuity in metric (gen. 1st countable) : Letx_n \rightarrow x , y_n \rightarrow y$, we want to show$x_ny_n \rightarrow xy$: We have that$(x_n-x)(y_n-y) \rightarrow 0$as$n \rightarrow \infty$, since each of the terms on the left can be made small -enough. Now:$x_ny_n -xy= x_ny_n ...

0

No. The Gelfand topology is the same as the usual topology on $\Bbb T$. The norm topology is discrete: Say $z,w\in\Bbb T$ and $z\ne w$. The dual of $L^1$ is $L^\infty$, so the norm of $z-w$, as an element of $L^1(\Bbb Z)^*$, is $$||z-w||=\sup_{n\in\Bbb Z}|z^n-w^n|=\sup_{n\in\Bbb Z}|1-(z/w)^n|.$$ It's easy to see that there exists $c>0$ such that if ...

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