This may be overkill, but the cantor function with $S=$ the middle-thirds cantor set will work:
2 properties of this function are: $S^\circ = \emptyset$ and $f(S)=[0,1]$
↧
Channel:
Counter example of continous function such that there is Set S with $f(S)^\circ \subset (f(S^\circ))$ - Mathematics Stack Exchange
X
Are you the publisher?
Claim or
contact us
about this channel.
X
0
Channel Details:
- Title: Counter example of continous function such that there is Set S with $f(S)^\circ \subset (f(S^\circ))$ - Mathematics Stack Exchange
- Channel Number: 78566829
- Language: English
- Registered On: March 15, 2024, 11:07 pm
- Number of Articles: 2
- Latest Snapshot: March 15, 2024, 11:08 pm
- RSS URL: https://math.stackexchange.com/feeds/question/3028720
- Publisher: https://math.stackexchange.com/q/3028720
- Description:
- Catalog: //continous50.rssing.com/catalog.php?indx=78566829
© 2025 //www.rssing.com