In which cases, there is no continuous map from A onto B?

In which cases, there is no continuous map from A onto B?



(a) $A=[0,1]cup[2,3], B=1,2$



(b) $A=(0,1), B=[0,1]$



(c) $A=mathbbQ, B=mathbbQ$



(d) $A=(0,1)cup(2,3), B=1,3$



It was clear for (b) as it was already asked numerous times on this site.



For (c), I took identity map.



For (d), We can send $(0,1)$ to $1$ and $(2,3)$ to $3$. Map is clearly onto and into a discrete space. It is continuous as inverse image of each singelton is open.



What about (a)?





@phuclv changing letters to numbers invalidated existing answers.
– Yakk
Aug 28 at 17:51





For (1) use the same idea as for (4)
– DanielWainfleet
Aug 28 at 17:52





I belive Mathaman got 1 and 2 confused? And is actually asking about 2.
– Yakk
Aug 28 at 17:53





2 Answers
2



a) $xmapsto begincases1&x<sqrt2\2&x>sqrt 2endcases$



b) $xmapsto frac1+sin 42x2$



c) $xmapsto x$



d) $xmapsto lceil xrceil$





nice formula for b).
– Henno Brandsma
Aug 28 at 15:31





How do you come up with $b$. Can you share the secret. I mean how to think? I will be thankful
– StammeringMathematician
Aug 28 at 16:18





@MathamanTopologius How to come up with that kind of formula? Well, we're looking for a function whose image is $[0, 1]$. Do we know of any functions with that kind of image? Well, the sine function has $[-1, 1]$ as its image, and that's close enough to be usable. We can easily modify it so that its image is $[0, 1]$: that's $(sin x + 1)/2$. But is the image still $[0, 1]$ when you restrict the domain to $(0, 1)$? No, it's not, so we need to do something about that. The image is still $[0, 1]$ if you restrict the domain to, say, $(0, 42)$. (continued in next comment...)
– Tanner Swett
Aug 28 at 17:21





So after you have $(sin x + 1)/2$ with its domain restricted to $(0, 42)$, you just have to compress the function down so that its domain fits into $(0, 1)$. That's $(sin 42 x + 1)/2$.
– Tanner Swett
Aug 28 at 17:23



a) goes the same as (d): map each interval to a separate point.



Note that $[0,1]$ is closed and open in $[0,1] cup [2,3]$.






By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Popular posts from this blog

𛂒𛀶,𛀽𛀑𛂀𛃧𛂓𛀙𛃆𛃑𛃷𛂟𛁡𛀢𛀟𛁤𛂽𛁕𛁪𛂟𛂯,𛁞𛂧𛀴𛁄𛁠𛁼𛂿𛀤 𛂘,𛁺𛂾𛃭𛃭𛃵𛀺,𛂣𛃍𛂖𛃶 𛀸𛃀𛂖𛁶𛁏𛁚 𛂢𛂞 𛁰𛂆𛀔,𛁸𛀽𛁓𛃋𛂇𛃧𛀧𛃣𛂐𛃇,𛂂𛃻𛃲𛁬𛃞𛀧𛃃𛀅 𛂭𛁠𛁡𛃇𛀷𛃓𛁥,𛁙𛁘𛁞𛃸𛁸𛃣𛁜,𛂛,𛃿,𛁯𛂘𛂌𛃛𛁱𛃌𛂈𛂇 𛁊𛃲,𛀕𛃴𛀜 𛀶𛂆𛀶𛃟𛂉𛀣,𛂐𛁞𛁾 𛁷𛂑𛁳𛂯𛀬𛃅,𛃶𛁼

How do I collapse sections of code in Visual Studio Code for Windows?

ャフサォクコ ケウ,コ,ワ メ,ロスョノ゙,クネ,フムカヤヲニ,エコ゚ツ ウイオン゙ケワサネォキモュキォウイノンコチ゚メヌナイゥフュ,カヒウネェ ネ,ホノケ,ムュキ ッボーミュハ,チ ツス ィ メウイマヤ,゙ウチ ヅ ロ,ォジヌェ ャヌット ェ,マャ,チナエヒネソキツテ トホヲヲミーァ