# Max/Min proof I can't follow

## Homework Statement

Give an example of a bounded set that has neither a maximum nor a minimum. (The proof below is given by the book).

We claim that the set ##(0,2)## is bounded and has neither a maximum nor a minimum.

Proof: For each ##x \epsilon (0,2)##, we know that ##0 < x < 2##. Therefore 0 is a lower bound of the set and 2 is an upper bound. Thus, (0,2) is bounded. To see that it has no maximum, suppose to the contrary that ##s## is a maximum of the set ##(0,2)##. Then, by definition of maximum, s must be in the set ##(0,2)##. But
##0 < s < \frac {2+s}{2} < 2## and therefore ##\frac {2+s}{2}## is in the set (0,2) and larger than s, a contradiction. In a similar fashion, you can check that there is no minimum.

## The Attempt at a Solution

I don't get where ##\frac {2+s}{2}## comes from. I know that since ##s < 2##, then
##s + 2 < 2 + 2## so ##s + 2 < 4## so ##\frac {s+2}{2} < 2##. But how do we know ## s < \frac {s+2}{2} ##

andrewkirk
Homework Helper
Gold Member
how do we know ## s < \frac {s+2}{2} ##
##s=\frac{s+s}2<\frac{s+2}2## because ##0<s<2##.

##s=\frac{s+s}2<\frac{s+2}2## because ##0<s<2##.
Thank you, now I get it

Ok, and to show minimum we would do this:

Suppose ##h## is a minimum of ##(0,2)##. Then ##0 < h < 2## by definition of minimum. But ##0 < \frac {h}{2} < h < 2##. Thus h is not a minimum, a contradiction. We conclude that ##(0,2)## does not have a minimum. []

note: for the minimum, we could have divided ##h## by any ##n > 0## and would have found another minimum.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Give an example of a bounded set that has neither a maximum nor a minimum. (The proof below is given by the book).

We claim that the set ##(0,2)## is bounded and has neither a maximum nor a minimum.

Proof: For each ##x \epsilon (0,2)##, we know that ##0 < x < 2##. Therefore 0 is a lower bound of the set and 2 is an upper bound. Thus, (0,2) is bounded. To see that it has no maximum, suppose to the contrary that ##s## is a maximum of the set ##(0,2)##. Then, by definition of maximum, s must be in the set ##(0,2)##. But
##0 < s < \frac {2+s}{2} < 2## and therefore ##\frac {2+s}{2}## is in the set (0,2) and larger than s, a contradiction. In a similar fashion, you can check that there is no minimum.

## The Attempt at a Solution

I don't get where ##\frac {2+s}{2}## comes from. I know that since ##s < 2##, then
##s + 2 < 2 + 2## so ##s + 2 < 4## so ##\frac {s+2}{2} < 2##. But how do we know ## s < \frac {s+2}{2} ##
{"ft_lang_label":"__label__en","ft_lang_prob":0.8555487,"math_prob":0.9397364,"size":975,"snap":"2021-31-2021-39","text_gpt3_token_len":330,"char_repetition_ratio":0.12976313,"word_repetition_ratio":0.030456852,"special_character_ratio":0.3979487,"punctuation_ratio":0.12,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99481565,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T10:11:21Z\",\"WARC-Record-ID\":\"<urn:uuid:9a87400d-1264-41c0-a663-bcee7a38ffa0>\",\"Content-Length\":\"81270\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aeb4ccff-623a-4f92-9713-eab78007e0bc>\",\"WARC-Concurrent-To\":\"<urn:uuid:d268de83-19bb-43fb-9715-d3e53c7579d9>\",\"WARC-IP-Address\":\"104.26.15.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/max-min-proof-i-cant-follow.927977/#post-5858048\",\"WARC-Payload-Digest\":\"sha1:F3CV2ADWADKXHPQ7IRNENGKTZR325D4C\",\"WARC-Block-Digest\":\"sha1:GMX6BUP7EUYD7OC7TQO5ZX5YF4IP42SI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057508.83_warc_CC-MAIN-20210924080328-20210924110328-00597.warc.gz\"}"}