Intereting Posts

A continuous bijection $f:\mathbb{R}\to \mathbb{R}$ is an homeomorphism?
Closed form formula for $\sum\limits_{k=1}^n k^k$
Is sin(x) necessarily irrational where x is rational?
Is it possible to define countability without referring the natural numbers?
What are “Lazard” sheaves?
On the spectrum of the sum of two commuting elements in a Banach algebra
Delta (Dirac) function integral
Evaluating $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$
Prove the integral of $f$ is positive if $f ≥ 0$, $f$ continuous at $x_0$ and $f(x_0)>0$
Prove that $f$ has a minimum
Why can a Venn diagram for 4+ sets not be constructed using circles?
Is $i = \sqrt{e^{\pi\sqrt{e^{\pi\sqrt\ldots}}}}$?
Showing that a limit exists and showing $f$ is not integrable.
Making Change for a Dollar (and other number partitioning problems)
line at infinity

Let $X$ and $Y$ metric spaces, $f$ is an injective from $X$ to $Y$, and $f$ sets every compact set in $X$ to compact set in $Y$. How to prove $f$ is continuous map?

Any comments and advice will be appreciated.

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- Illustration Proof that every sequence of real numbers has monotone subsequence

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- Is there an algebraic homomorphism between two Banach algebras which is not continuous?

Since $X$ and $Y$ are metric spaces, it suffices to show that if $\langle x_n:n\in\Bbb N\rangle$ is a convergent sequence in $X$ with limit $x$, then $\langle f(x_n):n\in\Bbb N\rangle$ is a convergent sequence in $Y$ with limit $f(x)$; in words, *f preserves convergent sequences*.

Suppose that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ in $X$. If there is an $n_0\in\Bbb N$ such that $x_n=x$ for all $n\ge n_0$, it’s trivially true that $\langle f(x_n):n\in\Bbb N\rangle\to f(x)$, so assume (by passing to a subsequence if necessary) that $\langle x_n:n\in\Bbb N\rangle$ is a sequence of distinct points. (Since $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ and is not eventually constant at $x$, it cannot have a constant infinite subsequence: for each $n\in\Bbb N$ there must be an $m>n$ such that $x_k\ne x_n$ whenever $k\ge m$.)

For each $n\in\Bbb N$ set $K_n=\{x\}\cup\{x_k:k\ge n\}$; each $K_n$ is compact and infinite. (Why?) By hypothesis, therefore, each $f[K_n]$ is compact.

For convenience let $y=f(x)$, and let $y_n=f(x_n)$ and $H_n=f[K_n]$ for $n\in\Bbb N$. By hypothesis each $H_n$ is compact and infinite, so each contains a limit point. Fix $n\in\Bbb N$. For each $k\ge n$, $Y\setminus H_{k+1}$ is an open nbhd of $y_k$ that contains only finitely many points of $H_n$ (why?), so $y_k$ can’t be a limit point of $H_n$. Thus, for each $n\in\Bbb N$ the only possible limit point of $H_n$ is $y$ itself. From here you should be able to prove without too much trouble that $\langle y_n:n\in\Bbb N\rangle\to y$ and hence that $f$ is continuous.

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