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A short proof that if $f$ is continuous then $f^ {-1}$ continuous
I learned a theorem that if $f$ is continuous and bijective then $f^ {-1}$ is continuous I went online to search for a proof and saw a really long proof in this link
Proof of Continuous compounding formula - Mathematics Stack Exchange
12 Following is the formula to calculate continuous compounding A = P e^(RT) Continuous Compound Interest Formula where, P = principal amount (initial investment) r = annual interest rate (as a decimal) t = number of years A = amount after time t The above is specific to continuous compounding
Showing that $\\ker T$ is closed if and only if $T$ is continuous.
Clearly if $f$ is continuous then its kernel is closed set for the converse, assume that $f\neq0$ and that $f^ {-1} (\ {0\})$ is a closed set Pick some $e$ in $X$ with $f (e)=1$
Show that the function $f (x, y)$ = $xy$ is continuous.
The projection $ (x,y)\mapsto x$ is a linear transformation and in finite dimensional space $\Bbb R^2$ it's continuous The same for the second projection and you know the rest of the story
If $f,g$ are continuous functions, then $fg$ is continuous?
I believe it follows from the fact that we showed $f+g$ is continuous whenever $f$ and $g$ are continuous Indeed, if $g$ is continuous, then $-g$ is clearly continuous