I don't want to give it away, but I'll give two hints: 1. By invariance of domain, a continuous injection from an open subset of R into R has to be a homeomorphism onto its image. 2. Try to use the sequence formulation of continuity to reach a contradiction.
That wasn't your original question. Anyway, a continuous injection defined on some interval is increasing. (Prove this yourself.) Show that f(x) = 1 is therefore impossible.
@HarryGindi I've posted this for others, who might not be familiar with using MathJax in chat. Surely, for a chat-veteran like you, this is a familiar thing.
Problem: If $J$ is uncountable, then $\Bbb{R}^J$ is not metrizable. Attempt: If $\Bbb{R}^J$ were metrizable, then the subspace $X := [0,1]^J$ would also be metrizable. In fact, it would be a compact metric space by Tychonoff's theorem. If so, then it is limit point compact and the set $A := \{e_i \mid i \in J\}$ has a limit point in $X$, call it $(x_i)$....
Here's where things breakdown. My strategy was to show that $A$ cannot in fact have a limit point. I was able to show that $(x_i)$ cannot be in $A$, but unfortunately it seems that $0$ is a limit point of $A$. Is there any way of salvaging the proof?
For the next transition table:
$$\begin{array}{|c|c|c|c|}\hline&0&1&2\\\hline a&a&b&d\\\hline b&a&b&c\\\hline c&c&d&a\\\hline d&c&c&a\\\hline\end{array}\\a\text{ is initial state}\\\{a,d\}\text{ are final states}$$
Make the digraph of the finite automaton and indicate if it is DFA (determinist...
Definition of compact in Real Analysis, Carothers, 1ed said that:
In Example 8.1 (c), he claimed that closed ball $\{x: \|x\| \leq 1\}$ in $l^{\infty}$ is not compact. Why?
See GA316's answer. In it, he makes the following claim: "Actually you can prove stronger result that, they are discrete. But the only discrete subsets of a compact sets are finite." Isn't this false? The set $\{\frac{1}{n} \mid n \in \Bbb{N}\}$ is an infinite discrete subspace of the compact space $[0,1]$, right?
If he/she is saying what I think he/she is, I believe it is false...right?
@LeakyNun first of all, the notation $(r)$ doesn't make sense in the noncommutative setting, so let's use $R/Rr$ instead. You're right that $M[r]=\{x \in M \mid r \cdot x =0\}$ isn't necessarily a submodule. But it is still true that $\mathrm{Hom}_R(R/Rr,M)=M[r]$. In general there is no reason to expect that $\mathrm{Hom}_R(M,N)$ is a $R$-module for two (say left) $R$-modules $M,N$
$\newcommand{Ext}{\operatorname{Ext}}\newcommand{Hom}{\operatorname{Hom}_R}$Let $R$ be a commutative ring with unity and $r \in R$.
Applying snake lemma to the following diagram:
$$\begin{array}{c} 0 & \longrightarrow & A~~ & \longrightarrow & B~~ & \longrightarrow & C~~ & \longrightarrow & 0 \\ &
@LeakyNun about your question: if $r$ is not a zero divisor, then your six-term exact sequence is just the exact sequence $0 \to \mathrm{Tor}_R^1(R/rR,A) \to \mathrm{Tor}_R^1(R/rR,B) \to \mathrm{Tor}_R^1(R/rR,C) \to R/rR \otimes_R A \to R/rR \otimes_R B \to R/rR \otimes_R C \to 0$
What's the easiest way to show that $x \mapsto (\cos (2 \pi x), \sin (2 \pi x))$ is a quotient map from $\Bbb{R}$ to $S^1$? Do it directly or show that it is a covering map, and then show every covering map is open?
I have a few equations that needs solving, but I'm scratching my head because it involves rounding ups and rounding downs. Plus, it's been like a decade ago since I've done this kind of thing.
I have realized that there ain't good libraries online to solve any equation given and seeing that I'm ...
@user193319 overkill approach: the map is a surjective continuous group homomorphism from a locally compact $\sigma$-compact Hausdorff group to a locally compact Hausdorff group, hence it is open
Can somebody pls tell me why this question has been downvoted? Am I missing something, or do I need to elaborate? https://math.stackexchange.com/q/3053372/333955
more generally if $H$ is a (closed, for continuous cohomology) normal subgroup in $G$ and $A$ is a $G$-module, then $A^H$ is a $G/H$-module and inflation is a map $H^\bullet(G/H,A^H) \to H^\bullet(G,A)$
but the situation with $\mathrm{Br}(K)$ and $\mathrm{Br}(L)$ is different. we have $L^{sep}=K^{sep}/L/K$ and the restriction map $H^2(G(K^{sep}/K),K^{sep})=\mathrm{Br}(K) \to H^2(G(L^{sep}/L),L^{sep})=\mathrm{Br}(L)$
For the next transition table:
$$\begin{array}{|c|c|c|c|}\hline&0&1&2\\\hline a&a&b&d\\\hline b&a&b&c\\\hline c&c&d&a\\\hline d&c&c&a\\\hline\end{array}\\a\text{ is initial state}\\\{a,d\}\text{ are final states}$$
Make the digraph of the finite automaton and indicate if it is DFA (determinist...
Definition of compact in Real Analysis, Carothers, 1ed said that:
In Example 8.1 (c), he claimed that closed ball $\{x: \|x\| \leq 1\}$ in $l^{\infty}$ is not compact. Why?
$\newcommand{Ext}{\operatorname{Ext}}\newcommand{Hom}{\operatorname{Hom}_R}$Let $R$ be a commutative ring with unity and $r \in R$.
Applying snake lemma to the following diagram:
$$\begin{array}{c} 0 & \longrightarrow & A~~ & \longrightarrow & B~~ & \longrightarrow & C~~ & \longrightarrow & 0 \\ &
